Solve Using the Quadratic Formula r^2-4r=12

$r^{2} - 4 r = 12$

Step 1

Subtract $12$ from both sides of the equation.

$r^{2} - 4 r - 12 = 0$

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 $a = 1$ , $b = - 4$ , and $c = - 12$ into the quadratic formula and solve for $r$ .

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

Step 4

Simplify.

Simplify the numerator.

Raise $- 4$ to the power of $2$ .

$r = \frac{4 \pm \sqrt{16 - 4 \cdot 1 \cdot - 12}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 12$ .

Multiply $- 4$ by $1$ .

$r = \frac{4 \pm \sqrt{16 - 4 \cdot - 12}}{2 \cdot 1}$

Multiply $- 4$ by $- 12$ .

$r = \frac{4 \pm \sqrt{16 + 48}}{2 \cdot 1}$

Add $16$ and $48$ .

$r = \frac{4 \pm \sqrt{64}}{2 \cdot 1}$

Rewrite $64$ as $8^{2}$ .

$r = \frac{4 \pm \sqrt{8^{2}}}{2 \cdot 1}$

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

$r = \frac{4 \pm 8}{2 \cdot 1}$

Multiply $2$ by $1$ .

$r = \frac{4 \pm 8}{2}$

Simplify $\frac{4 \pm 8}{2}$ .

$r = 2 \pm 4$

Step 5

The final answer is the combination of both solutions.

$r = 6, - 2$