Find the Roots (Zeros) 8x^2+12=7x^2+7x

$8 x^{2} + 12 = 7 x^{2} + 7 x$

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$7 x^{2} + 7 x = 8 x^{2} + 12$

Move all terms containing $x$ to the left side of the equation.

Subtract $8 x^{2}$ from both sides of the equation.

$7 x^{2} + 7 x - 8 x^{2} = 12$

Subtract $8 x^{2}$ from $7 x^{2}$ .

$- x^{2} + 7 x = 12$

Subtract $12$ from both sides of the equation.

$- x^{2} + 7 x - 12 = 0$

Factor the left side of the equation.

Factor $- 1$ out of $- x^{2} + 7 x - 12$ .

Factor $- 1$ out of $- x^{2}$ .

$- (x^{2}) + 7 x - 12 = 0$

Factor $- 1$ out of $7 x$ .

$- (x^{2}) - (- 7 x) - 12 = 0$

Rewrite $- 12$ as $- 1 (12)$ .

$- (x^{2}) - (- 7 x) - 1 \cdot 12 = 0$

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

$- (x^{2} - 7 x) - 1 \cdot 12 = 0$

Factor $- 1$ out of $- (x^{2} - 7 x) - 1 (12)$ .

$- (x^{2} - 7 x + 12) = 0$

Factor.

Factor $x^{2} - 7 x + 12$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $12$ and whose sum is $- 7$ .

$- 4, - 3$

Write the factored form using these integers.

$- ((x - 4) (x - 3)) = 0$

Remove unnecessary parentheses.

$- (x - 4) (x - 3) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 4 = 0$

$x - 3 = 0$

Set $x - 4$ equal to $0$ and solve for $x$ .

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

The final solution is all the values that make $- (x - 4) (x - 3) = 0$ true.

$x = 4, 3$

Find the Roots (Zeros) 8x^2+12=7x^2+7x

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