The Quadratic Formula

The quadratic formula is a single formula that solves every quadratic equation — no factoring required, no completing the square required. It works whether the solutions are integers, fractions, irrational numbers, or even complex numbers. If you learn one universal method for solving quadratics, this is it.

The formula is derived from completing the square on the general equation $ax^2 + bx + c = 0$, so the technique you learned in the previous section is the mathematical foundation for what follows here.

The Formula

For any quadratic equation in standard form $ax^2 + bx + c = 0$ (where $a \neq 0$):

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

This formula produces two solutions (because of the $\pm$):

$x_1 = \frac{-b + \sqrt{b^2 - 4ac}}{2a} \qquad x_2 = \frac{-b - \sqrt{b^2 - 4ac}}{2a}$

The expression under the square root, $b^2 - 4ac$, is called the discriminant. It determines the nature of the solutions — you will study it in detail in the next section.

Step-by-Step Process

1. Write the equation in standard form: $ax^2 + bx + c = 0$.
2. Identify $a$, $b$, and $c$.
3. Substitute into the formula.
4. Simplify under the square root first (compute the discriminant).
5. Evaluate both solutions (the $+$ case and the $-$ case).
6. Simplify radicals and fractions if possible.

Examples with Integer Solutions

Example 1: $x^2 + 5x + 6 = 0$

Identify: $a = 1$, $b = 5$, $c = 6$.

Substitute:

$x = \frac{-5 \pm \sqrt{5^2 - 4(1)(6)}}{2(1)}$

Compute the discriminant:

$b^2 - 4ac = 25 - 24 = 1$

Solve:

$x = \frac{-5 \pm \sqrt{1}}{2} = \frac{-5 \pm 1}{2}$

$x = \frac{-5 + 1}{2} = \frac{-4}{2} = -2 \qquad x = \frac{-5 - 1}{2} = \frac{-6}{2} = -3$

Answer: $x = -2$ or $x = -3$

This equation also factors as $(x + 2)(x + 3) = 0$, confirming the result. The quadratic formula works even when factoring would have been quicker — it just takes a few more steps.

Example 2: $2x^2 - 7x + 3 = 0$

Identify: $a = 2$, $b = -7$, $c = 3$.

Substitute:

$x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(2)(3)}}{2(2)}$

$x = \frac{7 \pm \sqrt{49 - 24}}{4} = \frac{7 \pm \sqrt{25}}{4} = \frac{7 \pm 5}{4}$

$x = \frac{7 + 5}{4} = \frac{12}{4} = 3 \qquad x = \frac{7 - 5}{4} = \frac{2}{4} = \frac{1}{2}$

Answer: $x = 3$ or $x = \dfrac{1}{2}$

Examples with Irrational Solutions

Example 3: $x^2 - 4x + 1 = 0$

Identify: $a = 1$, $b = -4$, $c = 1$.

$x = \frac{4 \pm \sqrt{16 - 4}}{2} = \frac{4 \pm \sqrt{12}}{2}$

Simplify the radical: $\sqrt{12} = \sqrt{4 \cdot 3} = 2\sqrt{3}$.

$x = \frac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3}$

Answer: $x = 2 + \sqrt{3} \approx 3.732$ or $x = 2 - \sqrt{3} \approx 0.268$

Notice we simplified the fraction by dividing every term in the numerator by 2. Always check if you can reduce.

Example 4: $3x^2 + 2x - 4 = 0$

Identify: $a = 3$, $b = 2$, $c = -4$.

$x = \frac{-2 \pm \sqrt{4 - 4(3)(-4)}}{2(3)} = \frac{-2 \pm \sqrt{4 + 48}}{6} = \frac{-2 \pm \sqrt{52}}{6}$

Simplify: $\sqrt{52} = \sqrt{4 \cdot 13} = 2\sqrt{13}$.

$x = \frac{-2 \pm 2\sqrt{13}}{6} = \frac{-1 \pm \sqrt{13}}{3}$

Answer: $x = \dfrac{-1 + \sqrt{13}}{3} \approx 0.869$ or $x = \dfrac{-1 - \sqrt{13}}{3} \approx -1.535$

When the Equation Is Not in Standard Form

You must rearrange before identifying $a$, $b$, and $c$.

Example 5: $x^2 = 6x - 5$

Step 1 — Move all terms to one side:

$x^2 - 6x + 5 = 0$

Identify: $a = 1$, $b = -6$, $c = 5$.

$x = \frac{6 \pm \sqrt{36 - 20}}{2} = \frac{6 \pm \sqrt{16}}{2} = \frac{6 \pm 4}{2}$

$x = \frac{10}{2} = 5 \qquad x = \frac{2}{2} = 1$

Answer: $x = 5$ or $x = 1$

Example 6: $4x^2 + 9 = 12x$

Rearrange: $4x^2 - 12x + 9 = 0$.

Identify: $a = 4$, $b = -12$, $c = 9$.

$x = \frac{12 \pm \sqrt{144 - 144}}{8} = \frac{12 \pm \sqrt{0}}{8} = \frac{12}{8} = \frac{3}{2}$

Answer: $x = \dfrac{3}{2}$ (one repeated solution). The discriminant is zero, meaning the parabola touches the $x$-axis at exactly one point.

Choosing the Best Method

You now have four methods for solving quadratics. Here is when to use each:

The quadratic formula always works, but it can be slower than factoring for simple problems. Develop the judgment to pick the fastest approach for each equation.

Real-World Application: Electrician — Calculating Power in a Circuit

In electrical work, the power formula for a resistive circuit is:

$P = I^2 R$

where $P$ is power in watts, $I$ is current in amps, and $R$ is resistance in ohms. An electrician is designing a circuit where the total power must not exceed 1,800 watts and the resistance is 8 ohms. What is the maximum current?

Setting up the equation:

$I^2 \cdot 8 = 1800$

$I^2 = 225$

$I = 15 \text{ amps}$

Now consider a more complex scenario. The power consumed by two series components is modeled by:

$P = 2I^2 + 5I$

If the total power dissipated is 63 watts, find the current:

$2I^2 + 5I = 63$

$2I^2 + 5I - 63 = 0$

Identify: $a = 2$, $b = 5$, $c = -63$.

$I = \frac{-5 \pm \sqrt{25 + 504}}{4} = \frac{-5 \pm \sqrt{529}}{4} = \frac{-5 \pm 23}{4}$

$I = \frac{18}{4} = 4.5 \qquad \text{or} \qquad I = \frac{-28}{4} = -7$

Current cannot be negative in this context, so $I = 4.5$ amps.

Answer: The current is 4.5 amps. The electrician would verify this is within the ampacity rating of the conductor and the breaker size.

An HVAC Application: Sizing a Heat Exchanger

An HVAC engineer models the heat transfer rate $Q$ (in BTU/hr) through a heat exchanger as:

$Q = -0.5T^2 + 30T + 200$

where $T$ is the temperature difference in degrees Fahrenheit between the hot and cold fluids. At what temperature difference does the heat transfer rate reach 500 BTU/hr?

$-0.5T^2 + 30T + 200 = 500$

$-0.5T^2 + 30T - 300 = 0$

Multiply by $-2$ to clear the decimal:

$T^2 - 60T + 600 = 0$

Identify: $a = 1$, $b = -60$, $c = 600$.

$T = \frac{60 \pm \sqrt{3600 - 2400}}{2} = \frac{60 \pm \sqrt{1200}}{2}$

Simplify: $\sqrt{1200} = \sqrt{400 \cdot 3} = 20\sqrt{3} \approx 34.64$.

$T = \frac{60 \pm 20\sqrt{3}}{2} = 30 \pm 10\sqrt{3}$

$T \approx 30 + 17.32 = 47.32 \quad \text{or} \quad T \approx 30 - 17.32 = 12.68$

Answer: The heat transfer rate reaches 500 BTU/hr at temperature differences of approximately 12.7 degrees F and 47.3 degrees F. The technician would operate the system near the lower value for energy efficiency.

Common Mistakes to Avoid

1. Getting the sign of $b$ wrong. In the formula, the first term in the numerator is $-b$. If $b = -7$, then $-b = 7$ (positive). If $b = 5$, then $-b = -5$. Be meticulous with the negative sign.

2. Forgetting to write the equation in standard form. The values of $a$, $b$, and $c$ come from $ax^2 + bx + c = 0$. If your equation is $x^2 = 3x + 10$, you must rearrange to $x^2 - 3x - 10 = 0$ first.

3. Computing $b^2 - 4ac$ incorrectly. This is the most error-prone step. Remember that $(-7)^2 = 49$ (not $-49$), and $-4(2)(-3) = +24$ (two negatives make a positive). Write out the arithmetic carefully.

4. Putting the $\pm$ only on part of the numerator. The formula is $\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. The entire numerator is divided by $2a$, not just the square root term. Use parentheses to keep everything together.

5. Not simplifying the final answer. Always reduce fractions and simplify radicals. The answer $\dfrac{4 \pm 2\sqrt{3}}{2}$ should be simplified to $2 \pm \sqrt{3}$.

Practice Problems

Test your understanding with these problems. Click to reveal each answer.

Key Takeaways

• The quadratic formula $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ solves any quadratic equation in standard form
• Always write the equation as $ax^2 + bx + c = 0$ first, then read off $a$, $b$, and $c$ — be careful with signs
• The discriminant $b^2 - 4ac$ tells you the nature of the solutions before you finish calculating
• Simplify your answer: reduce fractions and simplify radicals like $\sqrt{12} = 2\sqrt{3}$
• Factoring is faster when it works, but the quadratic formula is the guaranteed universal method
• In applied problems, check whether negative solutions make physical sense — often only the positive root applies

Return to Algebra 1 for more topics in this section.