Solve Using the Quadratic Formula x^2+36=12x

$x^{2} + 36 = 12 x$

Step 1

Subtract $12 x$ from both sides of the equation.

$x^{2} + 36 - 12 x = 0$

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 $a = 1$ , $b = - 12$ , and $c = 36$ into the quadratic formula and solve for $x$ .

$\frac{12 \pm \sqrt{{(- 12)}^{2} - 4 \cdot (1 \cdot 36)}}{2 \cdot 1}$

Step 4

Simplify.

Simplify the numerator.

Raise $- 12$ to the power of $2$ .

$x = \frac{12 \pm \sqrt{144 - 4 \cdot 1 \cdot 36}}{2 \cdot 1}$

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

Multiply $- 4$ by $1$ .

$x = \frac{12 \pm \sqrt{144 - 4 \cdot 36}}{2 \cdot 1}$

Multiply $- 4$ by $36$ .

$x = \frac{12 \pm \sqrt{144 - 144}}{2 \cdot 1}$

Subtract $144$ from $144$ .

$x = \frac{12 \pm \sqrt{0}}{2 \cdot 1}$

Rewrite $0$ as $0^{2}$ .

$x = \frac{12 \pm \sqrt{0^{2}}}{2 \cdot 1}$

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

$x = \frac{12 \pm 0}{2 \cdot 1}$

$12$ plus or minus $0$ is $12$ .

$x = \frac{12}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{12}{2}$

Divide $12$ by $2$ .

$x = 6$

Step 5

The final answer is the combination of both solutions.

$x = 6$ Double roots