Solving Quadratic Equations with the Quadratic Formula

Hello! Last lesson, you used function notation to describe a relationship and chose a domain that made sense in context. This time, the task is to find inputs that give a particular output: solve a quadratic equation using the quadratic formula. You will learn a reliable method for equations that are difficult to factor, see what its answers mean on a graph, and decide when a solution is relevant to a model.

What does solving a quadratic mean?

Consider the function $f (x) = x^{2} - 2 x - 3$ . Solving $x^{2} - 2 x - 3 = 0$ means finding every value of $x$ for which $f (x) = 0$ . On a graph, these values are the $x$ -intercepts, also called roots or zeros.

The graph of $y=x^2-2x-3$ meets the $x$-axis at $(-1,0)$ and $(3,0)$. Thus $x=-1$ and $x=3$ solve $x^2-2x-3=0$.

The graph makes the answers visible, but not every intercept lands neatly on an integer. A formula lets you calculate solutions even when a graph would give only estimates.

Any quadratic equation can be written in standard form

a x^{2} + b x + c = 0, a \neq = 0.

Here $a$ , $b$ , and $c$ are numbers, including their signs. The condition $a \neq = 0$ matters: without an $x^{2}$ term, the equation is not quadratic. Once the equation is in standard form, its solutions are given by

x = \frac{- b \pm b ^{2} - 4 a c}{2 a} .

The symbol $\pm$ instructs you to calculate both possibilities, one using addition and one using subtraction. The entire numerator is divided by $2 a$ .

How To Solve Quadratic Equations Using The Quadratic Formula

Watch “How To Solve Quadratic Equations Using The Quadratic Formula” by The Organic Chemistry Tutor for a complete substitution-and-check example, followed by one with a negative coefficient.

Watch the setup for standard form and the formula. In the first solution, follow how the plus and minus cases become two answers and how substitution checks one of them. Then watch the second example, paying particular attention to the sign when the coefficient of x is negative.

Apply the formula without losing a sign

Return to the equation in the graph:

x^{2} - 2 x - 3 = 0.

It is already in standard form. Its coefficients are $a = 1$ , $b = - 2$ , and $c = - 3$ . Substitute those signed values before simplifying:

x = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 ( 1 ) ( - 3 )}{2 ( 1 )} = \frac{2 \pm 4 + 12}{2} = \frac{2 \pm 4}{2} .

Now evaluate the two cases separately:

x = \frac{2 + 4}{2} = 3 or x = \frac{2 - 4}{2} = - 1.

These agree with the graph. Substitution confirms them without relying on the drawing: $3^{2} - 2 (3) - 3 = 0$ , and $(- 1)^{2} - 2 (- 1) - 3 = 0$ .

Two details prevent many errors. First, if $b = - 2$ , then $- b = 2$ and $b^{2} = (- 2)^{2} = 4$ . Parentheses make both signs clear. Second, keep the numerator together: $(2 \pm 4) /2$ does not mean $2 \pm (4/2)$ .

Rearrange first; keep exact answers when needed

The formula requires zero on one side of the equation. For example, to solve

2 x^{2} - 3 x = 1,

first subtract $1$ from both sides:

2 x^{2} - 3 x - 1 = 0.

Now $a = 2$ , $b = - 3$ , and $c = - 1$ . The formula gives

x = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 ( 2 ) ( - 1 )}{2 ( 2 )} = \frac{3 \pm 17}{4} .

The square root of $17$ does not simplify to a whole number. The two exact solutions are therefore

x = \frac{3 + 17}{4} and x = \frac{3 - 17}{4} .

As decimals, they are approximately $1.7808$ and $- 0.2808$ . An approximate answer may be useful in context, but retain the exact form unless rounding is requested: a rounded value will not usually satisfy the original equation exactly.

Solving by quadratic formula - Higher

Read the worked example in BBC Bitesize’s “Solving by quadratic formula - Higher” to see another equation whose answers require decimal approximations.

Under “Example,” locate the equation beginning x^2+6x-12=0. Starting with the sentence “First, identify the value of,” read the substitution, then continue through both decimal answers, stopping before “Question.” Notice that the expression containing \sqrt{84} is kept until the final calculation, and that the two answers are rounded to three significant figures.

A dependable written solution shows the rearranged equation, the signed coefficients, their substitution into the formula, and both resulting values. That record also makes a calculator or sign error much easier to find.

What the expression under the square root tells you

The quantity $b^{2} - 4 a c$ is called the discriminant. It determines how many distinct real solutions the equation has:

If $b^{2} - 4 a c > 0$ , the formula gives two distinct real solutions.
If $b^{2} - 4 a c = 0$ , the plus and minus cases coincide, giving one distinct real solution.
If $b^{2} - 4 a c < 0$ , there are no real solutions, because the formula would require the square root of a negative number.

For instance, $x^{2} - 6 x + 9 = 0$ has discriminant $36 - 36 = 0$ . Both cases in the formula give $x = 3$ ; the graph touches the $x$ -axis at one point. By contrast, $x^{2} + 2 x + 5 = 0$ has discriminant $4 - 20 = - 16$ , so it has no real $x$ -intercepts.

How to use the quadratic formula | Polynomial and rational functions | Algebra II | Khan Academy

Watch the no-real-solutions example in Khan Academy’s “How to use the quadratic formula” to connect a negative discriminant with the graph of a quadratic.

In the negative-discriminant example, watch the presenter first rearrange an equation into standard form, then obtain a negative value under the square root. Focus on the graph near the end: it does not meet the x-axis, consistent with the absence of real solutions.

Finally, recall the distinction from the previous lesson between a mathematical solution and an input allowed by a model. If $h (t) = - t^{2} + 4 t + 5$ models height in metres at time $t$ seconds, solving $h (t) = 0$ gives $t = - 1$ and $t = 5$ . Both solve the equation, but if the model starts at $t = 0$ , only 5 seconds is an admissible time. Always check the domain after doing the algebra.

Conclusion

To solve a quadratic with the formula, first write it as $a x^{2} + b x + c = 0$ . Identify the coefficients with their signs, substitute carefully, and evaluate both the plus and minus cases. Keep exact roots when appropriate, check your results, and use the discriminant to recognise whether there are two, one, or no distinct real solutions.

Today you found where a quadratic model equals zero. Next, you will use a quadratic model to find a maximum or minimum—a different feature of the same curve—and interpret it in a practical context.

Recommended resources

What does solving a quadratic mean?

Apply the formula without losing a sign

Rearrange first; keep exact answers when needed

What the expression under the square root tells you

Conclusion