Solve by Factoring 9x^2-8=6x

$9 x^{2} - 8 = 6 x$

Subtract $6 x$ from both sides of the equation.

$9 x^{2} - 8 - 6 x = 0$

Factor by grouping.

Reorder terms.

$9 x^{2} - 6 x - 8 = 0$

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 $a \cdot c = 9 \cdot - 8 = - 72$ and whose sum is $b = - 6$ .

Factor $- 6$ out of $- 6 x$ .

$9 x^{2} - 6 x - 8 = 0$

Rewrite $- 6$ as $6$ plus $- 12$

$9 x^{2} + (6 - 12) x - 8 = 0$

Apply the distributive property.

$9 x^{2} + 6 x - 12 x - 8 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(9 x^{2} + 6 x) - 12 x - 8 = 0$

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

$3 x (3 x + 2) - 4 (3 x + 2) = 0$

Factor the polynomial by factoring out the greatest common factor, $3 x + 2$ .

$(3 x + 2) (3 x - 4) = 0$

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

$3 x + 2 = 0$

$3 x - 4 = 0$

Set $3 x + 2$ equal to $0$ and solve for $x$ .

Set $3 x + 2$ equal to $0$ .

$3 x + 2 = 0$

Solve $3 x + 2 = 0$ for $x$ .

Subtract $2$ from both sides of the equation.

$3 x = - 2$

Divide each term in $3 x = - 2$ by $3$ and simplify.

Divide each term in $3 x = - 2$ by $3$ .

$\frac{3 x}{3} = \frac{- 2}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{- 2}{3}$

Divide $x$ by $1$ .

$x = \frac{- 2}{3}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{2}{3}$

Set $3 x - 4$ equal to $0$ and solve for $x$ .

Set $3 x - 4$ equal to $0$ .

$3 x - 4 = 0$

Solve $3 x - 4 = 0$ for $x$ .

Add $4$ to both sides of the equation.

$3 x = 4$

Divide each term in $3 x = 4$ by $3$ and simplify.

Divide each term in $3 x = 4$ by $3$ .

$\frac{3 x}{3} = \frac{4}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{4}{3}$

Divide $x$ by $1$ .

$x = \frac{4}{3}$

The final solution is all the values that make $(3 x + 2) (3 x - 4) = 0$ true.

$x = - \frac{2}{3}, \frac{4}{3}$

Solve by Factoring 9x^2-8=6x

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