Examples

x^{2} - 5 x + 6 = 0

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

Step 1

Use the quadratic formula to find the solutions.

\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 2

Substitute the values

a = 1

$a = 1$ ,

b = - 5

$b = - 5$ , and

c = 6

$c = 6$ into the quadratic formula and solve for

x

$x$ .

\frac{5 \pm \sqrt{{(- 5)}^{2} - 4 \cdot (1 \cdot 6)}}{2 \cdot 1}

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

Step 3

Simplify.

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Step 3.1

Simplify the numerator.

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Step 3.1.1

Raise

- 5

$- 5$ to the power of

2

$2$ .

x = \frac{5 \pm \sqrt{25 - 4 \cdot 1 \cdot 6}}{2 \cdot 1}

$x = \frac{5 \pm \sqrt{25 - 4 \cdot 1 \cdot 6}}{2 \cdot 1}$

Step 3.1.2

Multiply

- 4 \cdot 1 \cdot 6

$- 4 \cdot 1 \cdot 6$ .

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Step 3.1.2.1

Multiply

- 4

$- 4$ by

1

$1$ .

x = \frac{5 \pm \sqrt{25 - 4 \cdot 6}}{2 \cdot 1}

$x = \frac{5 \pm \sqrt{25 - 4 \cdot 6}}{2 \cdot 1}$

Step 3.1.2.2

Multiply

- 4

$- 4$ by

6

$6$ .

x = \frac{5 \pm \sqrt{25 - 24}}{2 \cdot 1}

$x = \frac{5 \pm \sqrt{25 - 24}}{2 \cdot 1}$

x = \frac{5 \pm \sqrt{25 - 24}}{2 \cdot 1}

$x = \frac{5 \pm \sqrt{25 - 24}}{2 \cdot 1}$

Step 3.1.3

Subtract

24

$24$ from

25

$25$ .

x = \frac{5 \pm \sqrt{1}}{2 \cdot 1}

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

Step 3.1.4

Any root of

1

$1$ is

1

$1$ .

x = \frac{5 \pm 1}{2 \cdot 1}

$x = \frac{5 \pm 1}{2 \cdot 1}$

x = \frac{5 \pm 1}{2 \cdot 1}

$x = \frac{5 \pm 1}{2 \cdot 1}$

Step 3.2

Multiply

2

$2$ by

1

$1$ .

x = \frac{5 \pm 1}{2}

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

x = \frac{5 \pm 1}{2}

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

Step 4

The final answer is the combination of both solutions.

x = 3, 2

$x = 3, 2$

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