Find the Roots (Zeros) 9x^2-4=7x^2-7x

$9 x^{2} - 4 = 7 x^{2} - 7 x$

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

$7 x^{2} - 7 x = 9 x^{2} - 4$

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

Subtract $9 x^{2}$ from both sides of the equation.

$7 x^{2} - 7 x - 9 x^{2} = - 4$

Subtract $9 x^{2}$ from $7 x^{2}$ .

$- 2 x^{2} - 7 x = - 4$

Add $4$ to both sides of the equation.

$- 2 x^{2} - 7 x + 4 = 0$

Factor the left side of the equation.

Factor $- 1$ out of $- 2 x^{2} - 7 x + 4$ .

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

$- (2 x^{2}) - 7 x + 4 = 0$

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

$- (2 x^{2}) - (7 x) + 4 = 0$

Rewrite $4$ as $- 1 (- 4)$ .

$- (2 x^{2}) - (7 x) - 1 \cdot - 4 = 0$

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

$- (2 x^{2} + 7 x) - 1 \cdot - 4 = 0$

Factor $- 1$ out of $- (2 x^{2} + 7 x) - 1 (- 4)$ .

$- (2 x^{2} + 7 x - 4) = 0$

Factor.

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot - 4 = - 8$ and whose sum is $b = 7$ .

Factor $7$ out of $7 x$ .

$- (2 x^{2} + 7 (x) - 4) = 0$

Rewrite $7$ as $- 1$ plus $8$

$- (2 x^{2} + (- 1 + 8) x - 4) = 0$

Apply the distributive property.

$- (2 x^{2} - 1 x + 8 x - 4) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$- ((2 x^{2} - 1 x) + 8 x - 4) = 0$

Factor out the greatest common factor (GCF) from each group.

$- (x (2 x - 1) + 4 (2 x - 1)) = 0$

Factor the polynomial by factoring out the greatest common factor, $2 x - 1$ .

$- ((2 x - 1) (x + 4)) = 0$

Remove unnecessary parentheses.

$- (2 x - 1) (x + 4) = 0$

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

$2 x - 1 = 0$

$x + 4 = 0$

Set $2 x - 1$ equal to $0$ and solve for $x$ .

Set $2 x - 1$ equal to $0$ .

$2 x - 1 = 0$

Solve $2 x - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$2 x = 1$

Divide each term in $2 x = 1$ by $2$ and simplify.

Divide each term in $2 x = 1$ by $2$ .

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

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

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

The final solution is all the values that make $- (2 x - 1) (x + 4) = 0$ true.

$x = \frac{1}{2}, - 4$

Find the Roots (Zeros) 9x^2-4=7x^2-7x

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