Solve Using the Quadratic Formula 7r^2-14r=-7

$7 r^{2} - 14 r = - 7$

Step 1

Add $7$ to both sides of the equation.

$7 r^{2} - 14 r + 7 = 0$

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 $a = 7$ , $b = - 14$ , and $c = 7$ into the quadratic formula and solve for $r$ .

$\frac{14 \pm \sqrt{{(- 14)}^{2} - 4 \cdot (7 \cdot 7)}}{2 \cdot 7}$

Step 4

Simplify.

Simplify the numerator.

Raise $- 14$ to the power of $2$ .

$r = \frac{14 \pm \sqrt{196 - 4 \cdot 7 \cdot 7}}{2 \cdot 7}$

Multiply $- 4 \cdot 7 \cdot 7$ .

Multiply $- 4$ by $7$ .

$r = \frac{14 \pm \sqrt{196 - 28 \cdot 7}}{2 \cdot 7}$

Multiply $- 28$ by $7$ .

$r = \frac{14 \pm \sqrt{196 - 196}}{2 \cdot 7}$

Subtract $196$ from $196$ .

$r = \frac{14 \pm \sqrt{0}}{2 \cdot 7}$

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

$r = \frac{14 \pm \sqrt{0^{2}}}{2 \cdot 7}$

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

$r = \frac{14 \pm 0}{2 \cdot 7}$

$14$ plus or minus $0$ is $14$ .

$r = \frac{14}{2 \cdot 7}$

Multiply $2$ by $7$ .

$r = \frac{14}{14}$

Divide $14$ by $14$ .

$r = 1$

Step 5

The final answer is the combination of both solutions.

$r = 1$ Double roots