Quadratic Formula Calculator

What is the quadratic formula?

The quadratic formula is the reliable way to solve any equation that fits the shape ax² + bx + c = 0. Some quadratics factor neatly in your head, but plenty of them do not, and that is exactly where this formula earns its keep. Plug in the three coefficients and it hands you both solutions every time, whether they turn out to be tidy whole numbers, messy decimals, or complex numbers.

Here is the formula itself:

where:

• a is the coefficient of x² (it cannot be zero, or the equation stops being quadratic)
• b is the coefficient of x
• c is the constant term
• the ± symbol means you work it out twice, once adding the square root and once subtracting it, which is why a quadratic usually has two solutions

The discriminant tells you what to expect

The part under the square root, b² − 4ac, is called the discriminant. Before you finish the whole calculation, it already tells you what kind of answers you are going to get:

• If it is positive, you get two different real solutions.
• If it is exactly zero, the two solutions land on the same value, so there is one repeated root.
• If it is negative, there are no real solutions, and the answers come out as a pair of complex numbers.

First find the discriminant: (−5)² − 4(1)(6) = 25 − 24 = 1. It is positive, so expect two real roots. Now finish the formula:

That gives x = 3 and x = 2. You can check both by substituting them back into the original equation, and sure enough, each one makes it equal zero.

How to use this calculator

1. Enter the values of \(a\), \(b\), and \(c\) from your equation.
2. Check that the equation is written in standard form, \(ax^2+bx+c=0\).
3. Use the formula to substitute your numbers into \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\).
4. Simplify to get the two possible values of \(x\), if they are real.

The formula explained

The quadratic formula computes the solutions of any quadratic equation in standard form. The expression under the square root, \(b^2-4ac\), tells you whether the equation has two real solutions, one real solution, or no real solutions.

• a = the coefficient of \(x^2\)
• b = the coefficient of \(x\)
• c = the constant term
• x = the unknown value that makes the equation true

Step by step method

1. First, rewrite the equation in standard form, so one side is zero.
2. Next, identify \(a\), \(b\), and \(c\), then substitute them into the quadratic formula.
3. Finally, simplify the square root and fraction to find the solution or solutions.

Worked example

Problem. Solve \(x^2-5x+6=0\) using the quadratic formula.

1. Here, \(a=1\), \(b=-5\), and \(c=6\).
2. Substitute into the formula: \(x=\frac{-(-5)\pm\sqrt{(-5)^2-4(1)(6)}}{2(1)}=\frac{5\pm\sqrt{25-24}}{2}=\frac{5\pm 1}{2}\).
3. So \(x=\frac{6}{2}=3\) or \(x=\frac{4}{2}=2\).

Tips and common mistakes

• Be careful with negative signs when substituting \(b\), especially if \(b\) is negative.
• If \(b^2-4ac<0\), the equation has no real solutions because the square root of a negative number is not real.

Frequently asked questions