Solve by Factoring 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

Factor $7$ out of $7 r^{2} - 14 r + 7$ .

Factor $7$ out of $7 r^{2}$ .

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

Factor $7$ out of $- 14 r$ .

$7 (r^{2}) + 7 (- 2 r) + 7 = 0$

Factor $7$ out of $7$ .

$7 (r^{2}) + 7 (- 2 r) + 7 (1) = 0$

Factor $7$ out of $7 (r^{2}) + 7 (- 2 r)$ .

$7 (r^{2} - 2 r) + 7 (1) = 0$

Factor $7$ out of $7 (r^{2} - 2 r) + 7 (1)$ .

$7 (r^{2} - 2 r + 1) = 0$

Step 3

Factor using the perfect square rule.

Rewrite $1$ as $1^{2}$ .

$7 (r^{2} - 2 r + 1^{2}) = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 r = 2 \cdot r \cdot 1$

Rewrite the polynomial.

$7 (r^{2} - 2 \cdot r \cdot 1 + 1^{2}) = 0$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = r$ and $b = 1$ .

$7 {(r - 1)}^{2} = 0$

Step 4

Divide each term in $7 {(r - 1)}^{2} = 0$ by $7$ and simplify.

Divide each term in $7 {(r - 1)}^{2} = 0$ by $7$ .

$\frac{7 {(r - 1)}^{2}}{7} = \frac{0}{7}$

Simplify the left side.

Cancel the common factor of $7$ .

Cancel the common factor.

$\frac{7 {(r - 1)}^{2}}{7} = \frac{0}{7}$

Divide ${(r - 1)}^{2}$ by $1$ .

${(r - 1)}^{2} = \frac{0}{7}$

Simplify the right side.

Divide $0$ by $7$ .

${(r - 1)}^{2} = 0$

Step 5

Set the $r - 1$ equal to $0$ .

$r - 1 = 0$

Step 6

Add $1$ to both sides of the equation.

$r = 1$