Solve Using the Quadratic Formula 0=x^2-15x+36

$0 = x^{2} - 15 x + 36$

Step 1

Move all the expressions to the left side of the equation.

Subtract $x^{2}$ from both sides of the equation.

$- x^{2} = - 15 x + 36$

Add $15 x$ to both sides of the equation.

$- x^{2} + 15 x = 36$

Subtract $36$ from both sides of the equation.

$- x^{2} + 15 x - 36 = 0$

Step 2

Use the quadratic formula to find the solutions.

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

Step 3

Substitute the values $a = - 1$ , $b = 15$ , and $c = - 36$ into the quadratic formula and solve for $x$ .

$\frac{- 15 \pm \sqrt{15^{2} - 4 \cdot (- 1 \cdot - 36)}}{2 \cdot - 1}$

Step 4

Simplify.

Simplify the numerator.

Raise $15$ to the power of $2$ .

$x = \frac{- 15 \pm \sqrt{225 - 4 \cdot - 1 \cdot - 36}}{2 \cdot - 1}$

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

Multiply $- 4$ by $- 1$ .

$x = \frac{- 15 \pm \sqrt{225 + 4 \cdot - 36}}{2 \cdot - 1}$

Multiply $4$ by $- 36$ .

$x = \frac{- 15 \pm \sqrt{225 - 144}}{2 \cdot - 1}$

Subtract $144$ from $225$ .

$x = \frac{- 15 \pm \sqrt{81}}{2 \cdot - 1}$

Rewrite $81$ as $9^{2}$ .

$x = \frac{- 15 \pm \sqrt{9^{2}}}{2 \cdot - 1}$

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

$x = \frac{- 15 \pm 9}{2 \cdot - 1}$

Multiply $2$ by $- 1$ .

$x = \frac{- 15 \pm 9}{- 2}$

Simplify $\frac{- 15 \pm 9}{- 2}$ .

$x = \frac{15 \pm 9}{2}$

Step 5

The final answer is the combination of both solutions.

$x = 12, 3$