Solve for x (16x^2-9)/(4x+3)=-6

$\frac{16 x^{2} - 9}{4 x + 3} = - 6$

Step 1

Factor each term.

Rewrite $16 x^{2}$ as ${(4 x)}^{2}$ .

$\frac{{(4 x)}^{2} - 9}{4 x + 3} = - 6$

Rewrite $9$ as $3^{2}$ .

$\frac{{(4 x)}^{2} - 3^{2}}{4 x + 3} = - 6$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 4 x$ and $b = 3$ .

$\frac{(4 x + 3) (4 x - 3)}{4 x + 3} = - 6$

Reduce the expression by cancelling the common factors.

Reduce the expression $\frac{(4 x + 3) (4 x - 3)}{4 x + 3}$ by cancelling the common factors.

Cancel the common factor.

$\frac{(4 x + 3) (4 x - 3)}{4 x + 3} = - 6$

Rewrite the expression.

$\frac{4 x - 3}{1} = - 6$

Divide $4 x - 3$ by $1$ .

$4 x - 3 = - 6$

Step 2

Solve the equation.

Move all terms not containing $x$ to the right side of the equation.

Add $3$ to both sides of the equation.

$4 x = - 6 + 3$

Add $- 6$ and $3$ .

$4 x = - 3$

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

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

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

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

Step 3

Exclude the solutions that do not make $\frac{16 x^{2} - 9}{4 x + 3} = - 6$ true.

$No solution$