Solve (3x+2)(3x+2)=25

$$(3x+2)(3x+2)=25$$

Solve for x

$$x=1$$

$$x = -\frac{7}{3} = -2\frac{1}{3} \approx -2.333333333$$

Steps Using the Quadratic Formula

Multiply 3x+2 and 3x+2 to get \left(3x+2\right)^{2}.

$$\left(3x+2\right)^{2}=25$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3x+2\right)^{2}.

$$9x^{2}+12x+4=25$$

Subtract 25 from both sides.

$$9x^{2}+12x+4-25=0$$

Subtract 25 from 4 to get -21.

$$9x^{2}+12x-21=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, 12 for b, and -21 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-12±\sqrt{12^{2}-4\times 9\left(-21\right)}}{2\times 9}$$

Square 12.

$$x=\frac{-12±\sqrt{144-4\times 9\left(-21\right)}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-12±\sqrt{144-36\left(-21\right)}}{2\times 9}$$

Multiply -36 times -21.

$$x=\frac{-12±\sqrt{144+756}}{2\times 9}$$

Add 144 to 756.

$$x=\frac{-12±\sqrt{900}}{2\times 9}$$

Take the square root of 900.

$$x=\frac{-12±30}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{-12±30}{18}$$

Now solve the equation x=\frac{-12±30}{18} when ± is plus. Add -12 to 30.

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

Divide 18 by 18.

$$x=1$$

Now solve the equation x=\frac{-12±30}{18} when ± is minus. Subtract 30 from -12.

$$x=-\frac{42}{18}$$

Reduce the fraction \frac{-42}{18} to lowest terms by extracting and canceling out 6.

$$x=-\frac{7}{3}$$

The equation is now solved.

$$x=1 x=-\frac{7}{3}$$

Steps for Completing the Square

Multiply 3x+2 and 3x+2 to get \left(3x+2\right)^{2}.

$$\left(3x+2\right)^{2}=25$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3x+2\right)^{2}.

$$9x^{2}+12x+4=25$$

Subtract 4 from both sides.

$$9x^{2}+12x=25-4$$

Subtract 4 from 25 to get 21.

$$9x^{2}+12x=21$$

Divide both sides by 9.

$$\frac{9x^{2}+12x}{9}=\frac{21}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$x^{2}+\frac{12}{9}x=\frac{21}{9}$$

Reduce the fraction \frac{12}{9} to lowest terms by extracting and canceling out 3.

$$x^{2}+\frac{4}{3}x=\frac{21}{9}$$

Reduce the fraction \frac{21}{9} to lowest terms by extracting and canceling out 3.

$$x^{2}+\frac{4}{3}x=\frac{7}{3}$$

Divide \frac{4}{3}, the coefficient of the x term, by 2 to get \frac{2}{3}. Then add the square of \frac{2}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{4}{3}x+\left(\frac{2}{3}\right)^{2}=\frac{7}{3}+\left(\frac{2}{3}\right)^{2}$$

Square \frac{2}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{4}{3}x+\frac{4}{9}=\frac{7}{3}+\frac{4}{9}$$

Add \frac{7}{3} to \frac{4}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{4}{3}x+\frac{4}{9}=\frac{25}{9}$$

Factor x^{2}+\frac{4}{3}x+\frac{4}{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}.

$$\left(x+\frac{2}{3}\right)^{2}=\frac{25}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{2}{3}\right)^{2}}=\sqrt{\frac{25}{9}}$$

Simplify.

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

Subtract \frac{2}{3} from both sides of the equation.

$$x=1 x=-\frac{7}{3}$$