How to Use the Quadratic Formula (With Examples)

Quadratic equations appear throughout algebra, physics, engineering, economics, and everyday problem-solving. Any time a quantity depends on the square of a variable — the height of a thrown ball, the area of a rectangle, the profit from pricing a product — you’re dealing with a quadratic relationship. The quadratic formula gives you a direct path to the solution every time.

This guide covers the formula itself, explains what the discriminant reveals before you even solve, walks through completing the square, introduces vertex form, and shows how quadratic equations model real-world situations. For instant solutions, use our Quadratic Formula Calculator.

The Quadratic Formula

A quadratic equation in standard form looks like this:

$ax^2 + bx + c = 0$

where a, b, and c are constants and a is not zero. The quadratic formula solves for x:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

The plus-or-minus sign means there are (potentially) two solutions: one using addition, one using subtraction.

Step-by-Step Process

1. Write the equation in standard form (everything on one side, zero on the other).
2. Identify a, b, and c.
3. Calculate the discriminant: b^2 - 4ac.
4. Plug a, b, and the discriminant into the formula.
5. Simplify to get your two solutions.
6. Verify by substituting each solution back into the original equation.

Worked Example 1

Solve: 2x^2 + 5x - 3 = 0

Identify coefficients: a = 2, b = 5, c = -3

Calculate the discriminant:

$b^2 - 4ac = 5^2 - 4(2)(-3) = 25 + 24 = 49$

Apply the formula:

$x = \frac{-5 \pm \sqrt{49}}{2(2)} = \frac{-5 \pm 7}{4}$

Two solutions:

• x = (-5 + 7) / 4 = 2/4 = 0.5
• x = (-5 - 7) / 4 = -12/4 = -3

Verify x = 0.5: 2(0.25) + 5(0.5) - 3 = 0.5 + 2.5 - 3 = 0. Correct.

Verify x = -3: 2(9) + 5(-3) - 3 = 18 - 15 - 3 = 0. Correct.

Worked Example 2

Solve: x^2 - 6x + 9 = 0

Identify: a = 1, b = -6, c = 9

Discriminant: (-6)^2 - 4(1)(9) = 36 - 36 = 0

Apply:

$x = \frac{6 \pm \sqrt{0}}{2} = \frac{6}{2} = 3$

There is only one solution: x = 3 (a repeated root). The parabola touches the x-axis at exactly one point.

The Discriminant: Your Preview of the Answer

The expression under the square root — b^2 - 4ac — is called the discriminant, often written as the Greek letter delta. It tells you the nature of the solutions before you finish calculating:

Discriminant Value	Number of Real Solutions	What It Means
Positive (> 0)	Two distinct real solutions	Parabola crosses x-axis at two points
Zero (= 0)	One repeated real solution	Parabola touches x-axis at its vertex
Negative (< 0)	No real solutions (two complex)	Parabola never reaches the x-axis

Example with a negative discriminant: Solve x^2 + 2x + 5 = 0

Discriminant: 2^2 - 4(1)(5) = 4 - 20 = -16

Since -16 is negative, there are no real solutions. The parabola y = x^2 + 2x + 5 sits entirely above the x-axis. In advanced mathematics, the solutions involve imaginary numbers: x = (-2 +/- 4i) / 2 = -1 +/- 2i.

Checking the discriminant first saves time. If you only need real-number answers and the discriminant is negative, you can stop immediately.

Completing the Square

The quadratic formula was originally derived by completing the square. Understanding this method deepens your grasp of why the formula works and gives you an alternative solving technique.

The Process

Starting with ax^2 + bx + c = 0:

1. Divide everything by a (so the x^2 coefficient becomes 1).
2. Move the constant to the other side.
3. Take half the x coefficient, square it, and add it to both sides.
4. Factor the left side as a perfect square.
5. Take the square root of both sides and solve.

Worked Example

Solve: x^2 + 8x + 5 = 0

Step 1: Already has a = 1, so skip division.

Step 2: Move constant: x^2 + 8x = -5

Step 3: Half of 8 is 4. Square it: 16. Add to both sides:

x^2 + 8x + 16 = -5 + 16 = 11

Step 4: Factor: (x + 4)^2 = 11

Step 5: Take the square root:

x + 4 = +/- sqrt(11)

x = -4 +/- sqrt(11)

• x = -4 + 3.317 = -0.683
• x = -4 - 3.317 = -7.317

This matches what the quadratic formula would give for a = 1, b = 8, c = 5.

Vertex Form and the Parabola

Every quadratic equation y = ax^2 + bx + c graphs as a parabola. Converting to vertex form reveals the parabola’s peak (or valley) directly.

Vertex form:

$y = a(x - h)^2 + k$

where (h, k) is the vertex of the parabola.

Finding the vertex from standard form:

$h = \frac{-b}{2a}, \quad k = f(h) = a \cdot h^2 + b \cdot h + c$

Example: Find the vertex of y = 3x^2 - 12x + 7.

• h = -(-12) / (2 x 3) = 12/6 = 2
• k = 3(4) - 12(2) + 7 = 12 - 24 + 7 = -5
• Vertex: (2, -5)
• Vertex form: y = 3(x - 2)^2 - 5

Since a = 3 (positive), the parabola opens upward, and the vertex is the minimum point. The lowest value of y is -5, occurring at x = 2.

Connection to the Roots

The vertex’s x-coordinate (h = -b / 2a) is exactly the midpoint of the two roots from the quadratic formula. If the roots are x = 1 and x = 3, the vertex sits at x = 2. The axis of symmetry of the parabola passes through the vertex.

Factoring: The Quick Alternative

When a quadratic factors neatly, factoring is faster than the formula. But many quadratics don’t factor into nice integers, which is when the formula becomes essential.

Example that factors: x^2 - 5x + 6 = 0

Look for two numbers that multiply to 6 and add to -5: that is -2 and -3.

(x - 2)(x - 3) = 0, so x = 2 or x = 3.

Example that doesn’t factor neatly: x^2 + 3x - 7 = 0

No integer pair multiplies to -7 and adds to 3. The quadratic formula gives:

$x = \frac{-3 \pm \sqrt{9 + 28}}{2} = \frac{-3 \pm \sqrt{37}}{2}$

x = (-3 + 6.083) / 2 = 1.541 or x = (-3 - 6.083) / 2 = -4.541

The quadratic formula works every time, whether the roots are integers, fractions, irrational, or complex.

Real-World Applications

Projectile Motion

When you throw a ball upward, its height follows a quadratic equation:

$h(t) = -\frac{1}{2}gt^2 + v_0 t + h_0$

where g is gravitational acceleration (9.8 m/s^2), v_0 is initial velocity, h_0 is initial height, and t is time.

Example: You throw a ball upward at 20 m/s from a height of 1.5 m. When does it hit the ground?

Set h(t) = 0: -4.9t^2 + 20t + 1.5 = 0

a = -4.9, b = 20, c = 1.5

Discriminant: 400 - 4(-4.9)(1.5) = 400 + 29.4 = 429.4

$t = \frac{-20 \pm \sqrt{429.4}}{-9.8}$

t = (-20 + 20.72) / -9.8 = -0.074 (negative time, discard)

t = (-20 - 20.72) / -9.8 = 4.155 seconds

The ball hits the ground after approximately 4.16 seconds.

Area Optimization

A farmer has 120 meters of fencing and wants to enclose the largest possible rectangular area against an existing wall (so fencing covers only three sides). If x is the width, the length is 120 - 2x, and the area is:

$A = x(120 - 2x) = -2x^2 + 120x$

This is a downward-opening parabola. Maximum area occurs at the vertex:

x = -120 / (2 x -2) = 30 meters

Area = 30 x (120 - 60) = 30 x 60 = 1,800 square meters

Revenue and Pricing

A store sells 200 units at $50 each. Market research shows that for every $5 price increase, they sell 10 fewer units. Revenue R as a function of the number of $5 increases (n) is:

$R = (50 + 5n)(200 - 10n) = -50n^2 + 500n + 10000$

Maximum revenue occurs at n = -500 / (2 x -50) = 5 increases. That means a price of $75 and sales of 150 units, giving revenue of $11,250 (compared to the original $10,000).

Solve any quadratic equation instantly with our Quadratic Formula Calculator.

Frequently Asked Questions

What if I forget the quadratic formula during a test?

You can always fall back on completing the square, which works for every quadratic equation and doesn’t require memorization of a separate formula. Alternatively, try factoring first — if the roots are integers, factoring is faster. A common mnemonic for the formula is singing it to the tune of “Pop Goes the Weasel”: “x equals negative b, plus or minus the square root, of b squared minus four a c, all over two a.”

Can a quadratic equation have more than two solutions?

No. A polynomial equation of degree n has at most n solutions (counting complex solutions and multiplicities). Since a quadratic is degree 2, it has exactly two solutions — though they might be the same number (a repeated root) or complex numbers (when the discriminant is negative). If you find three or more solutions, there’s an error in your algebra.

How do I know whether to use the quadratic formula or factoring?

If the coefficients are small integers and you can spot the factor pairs quickly, factoring is faster. If the coefficients are large, decimal, or the equation doesn’t factor neatly, use the formula. Many students check the discriminant first: if b^2 - 4ac is a perfect square (like 49, 144, or 0), the roots are rational and the equation likely factors. If the discriminant isn’t a perfect square, the roots are irrational and the formula is the only practical path.

What does it mean when there are no real solutions?

A negative discriminant means the parabola y = ax^2 + bx + c never crosses the x-axis. If a is positive, the entire parabola sits above the x-axis (the expression is always positive). If a is negative, it sits entirely below. In practical terms, this means the situation modeled by the equation has no valid answer. For instance, if a projectile motion equation yields no real solutions for when the ball reaches 500 meters, the ball simply never gets that high.

Can the quadratic formula solve equations that aren’t in standard form?

You must rearrange the equation into ax^2 + bx + c = 0 first. For example, 3x^2 = 7x - 2 becomes 3x^2 - 7x + 2 = 0 (subtract 7x and add 2 to both sides). Similarly, x(x + 4) = 12 expands to x^2 + 4x - 12 = 0. Once in standard form, identify a, b, and c and proceed with the formula as usual.