Solve Using the Quadratic Formula x^2-13x+36=0

$x^{2} - 13 x + 36 = 0$

Step 1

Use the quadratic formula to find the solutions.

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

Step 2

Substitute the values $a = 1$ , $b = - 13$ , and $c = 36$ into the quadratic formula and solve for $x$ .

$\frac{13 \pm \sqrt{{(- 13)}^{2} - 4 \cdot (1 \cdot 36)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 13$ to the power of $2$ .

$x = \frac{13 \pm \sqrt{169 - 4 \cdot 1 \cdot 36}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 36$ .

Multiply $- 4$ by $1$ .

$x = \frac{13 \pm \sqrt{169 - 4 \cdot 36}}{2 \cdot 1}$

Multiply $- 4$ by $36$ .

$x = \frac{13 \pm \sqrt{169 - 144}}{2 \cdot 1}$

Subtract $144$ from $169$ .

$x = \frac{13 \pm \sqrt{25}}{2 \cdot 1}$

Rewrite $25$ as $5^{2}$ .

$x = \frac{13 \pm \sqrt{5^{2}}}{2 \cdot 1}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \frac{13 \pm 5}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{13 \pm 5}{2}$

Step 4

The final answer is the combination of both solutions.

$x = 9, 4$