Solve (3x+2)(3x-4)=7

$$(3x+2)(3x-4)=7$$

Solve for x

$$x=-1$$

$$x = \frac{5}{3} = 1\frac{2}{3} \approx 1.666666667$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 3x+2 by 3x-4 and combine like terms.

$$9x^{2}-6x-8=7$$

Subtract 7 from both sides.

$$9x^{2}-6x-8-7=0$$

Subtract 7 from -8 to get -15.

$$9x^{2}-6x-15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, -6 for b, and -15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 9\left(-15\right)}}{2\times 9}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times 9\left(-15\right)}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-\left(-6\right)±\sqrt{36-36\left(-15\right)}}{2\times 9}$$

Multiply -36 times -15.

$$x=\frac{-\left(-6\right)±\sqrt{36+540}}{2\times 9}$$

Add 36 to 540.

$$x=\frac{-\left(-6\right)±\sqrt{576}}{2\times 9}$$

Take the square root of 576.

$$x=\frac{-\left(-6\right)±24}{2\times 9}$$

The opposite of -6 is 6.

$$x=\frac{6±24}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{6±24}{18}$$

Now solve the equation x=\frac{6±24}{18} when ± is plus. Add 6 to 24.

$$x=\frac{30}{18}$$

Reduce the fraction \frac{30}{18} to lowest terms by extracting and canceling out 6.

$$x=\frac{5}{3}$$

Now solve the equation x=\frac{6±24}{18} when ± is minus. Subtract 24 from 6.

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

Divide -18 by 18.

$$x=-1$$

The equation is now solved.

$$x=\frac{5}{3} x=-1$$

Steps for Completing the Square

Use the distributive property to multiply 3x+2 by 3x-4 and combine like terms.

$$9x^{2}-6x-8=7$$

Add 8 to both sides.

$$9x^{2}-6x=7+8$$

Add 7 and 8 to get 15.

$$9x^{2}-6x=15$$

Divide both sides by 9.

$$\frac{9x^{2}-6x}{9}=\frac{15}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$x^{2}+\left(-\frac{6}{9}\right)x=\frac{15}{9}$$

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

$$x^{2}-\frac{2}{3}x=\frac{15}{9}$$

Reduce the fraction \frac{15}{9} to lowest terms by extracting and canceling out 3.

$$x^{2}-\frac{2}{3}x=\frac{5}{3}$$

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.

$$x^{2}-\frac{2}{3}x+\left(-\frac{1}{3}\right)^{2}=\frac{5}{3}+\left(-\frac{1}{3}\right)^{2}$$

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

$$x^{2}-\frac{2}{3}x+\frac{1}{9}=\frac{5}{3}+\frac{1}{9}$$

Add \frac{5}{3} to \frac{1}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{2}{3}x+\frac{1}{9}=\frac{16}{9}$$

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}.

$$\left(x-\frac{1}{3}\right)^{2}=\frac{16}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{3}\right)^{2}}=\sqrt{\frac{16}{9}}$$

Simplify.

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

Add \frac{1}{3} to both sides of the equation.

$$x=\frac{5}{3} x=-1$$