Solve (3x-6)(3x+8)=-48

$$( 3 x - 6 ) ( 3 x + 8 ) = - 48$$

Solve for x

$$x=-\frac{2}{3}\approx -0.666666667$$

$$x=0$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 3x-6 by 3x+8 and combine like terms.

$$9x^{2}+6x-48=-48$$

Add 48 to both sides.

$$9x^{2}+6x-48+48=0$$

Add -48 and 48 to get 0.

$$9x^{2}+6x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, 6 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}}}{2\times 9}$$

Take the square root of 6^{2}.

$$x=\frac{-6±6}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{-6±6}{18}$$

Now solve the equation x=\frac{-6±6}{18} when ± is plus. Add -6 to 6.

$$x=\frac{0}{18}$$

Divide 0 by 18.

$$x=0$$

Now solve the equation x=\frac{-6±6}{18} when ± is minus. Subtract 6 from -6.

$$x=-\frac{12}{18}$$

Reduce the fraction \frac{-12}{18} to lowest terms by extracting and canceling out 6.

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

The equation is now solved.

$$x=0 x=-\frac{2}{3}$$

Steps for Completing the Square

Use the distributive property to multiply 3x-6 by 3x+8 and combine like terms.

$$9x^{2}+6x-48=-48$$

Add 48 to both sides.

$$9x^{2}+6x=-48+48$$

Add -48 and 48 to get 0.

$$9x^{2}+6x=0$$

Divide both sides by 9.

$$\frac{9x^{2}+6x}{9}=\frac{0}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$x^{2}+\frac{6}{9}x=\frac{0}{9}$$

Reduce the fraction \frac{6}{9} to lowest terms by extracting and canceling out 3.

$$x^{2}+\frac{2}{3}x=\frac{0}{9}$$

Divide 0 by 9.

$$x^{2}+\frac{2}{3}x=0$$

Divide \frac{2}{3}, the coefficient of the x term, by 2 to get \frac{1}{3}. Then add the square of \frac{1}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{2}{3}x+\left(\frac{1}{3}\right)^{2}=\left(\frac{1}{3}\right)^{2}$$

Square \frac{1}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=\frac{1}{9}$$

Factor x^{2}+\frac{2}{3}x+\frac{1}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{3}\right)^{2}=\frac{1}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{3}\right)^{2}}=\sqrt{\frac{1}{9}}$$

Simplify.

$$x+\frac{1}{3}=\frac{1}{3} x+\frac{1}{3}=-\frac{1}{3}$$

Subtract \frac{1}{3} from both sides of the equation.

$$x=0 x=-\frac{2}{3}$$