Solve (3x+2)^2+5(3x+2)-14=0

$$( 3 x + 2 ) ^ { 2 } + 5 ( 3 x + 2 ) - 14 = 0$$

Solve for x

$$x=-3$$

$$x=0$$

Steps Using Factoring

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

$$9x^{2}+12x+4+5\left(3x+2\right)-14=0$$

Use the distributive property to multiply 5 by 3x+2.

$$9x^{2}+12x+4+15x+10-14=0$$

Combine 12x and 15x to get 27x.

$$9x^{2}+27x+4+10-14=0$$

Add 4 and 10 to get 14.

$$9x^{2}+27x+14-14=0$$

Subtract 14 from 14 to get 0.

$$9x^{2}+27x=0$$

Factor out x.

$$x\left(9x+27\right)=0$$

To find equation solutions, solve x=0 and 9x+27=0.

$$x=0 x=-3$$

Steps Using the Quadratic Formula

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

$$9x^{2}+12x+4+5\left(3x+2\right)-14=0$$

Use the distributive property to multiply 5 by 3x+2.

$$9x^{2}+12x+4+15x+10-14=0$$

Combine 12x and 15x to get 27x.

$$9x^{2}+27x+4+10-14=0$$

Add 4 and 10 to get 14.

$$9x^{2}+27x+14-14=0$$

Subtract 14 from 14 to get 0.

$$9x^{2}+27x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, 27 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-27±\sqrt{27^{2}}}{2\times 9}$$

Take the square root of 27^{2}.

$$x=\frac{-27±27}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{-27±27}{18}$$

Now solve the equation x=\frac{-27±27}{18} when ± is plus. Add -27 to 27.

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

Divide 0 by 18.

$$x=0$$

Now solve the equation x=\frac{-27±27}{18} when ± is minus. Subtract 27 from -27.

$$x=-\frac{54}{18}$$

Divide -54 by 18.

$$x=-3$$

The equation is now solved.

$$x=0 x=-3$$

Steps for Completing the Square

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

$$9x^{2}+12x+4+5\left(3x+2\right)-14=0$$

Use the distributive property to multiply 5 by 3x+2.

$$9x^{2}+12x+4+15x+10-14=0$$

Combine 12x and 15x to get 27x.

$$9x^{2}+27x+4+10-14=0$$

Add 4 and 10 to get 14.

$$9x^{2}+27x+14-14=0$$

Subtract 14 from 14 to get 0.

$$9x^{2}+27x=0$$

Divide both sides by 9.

$$\frac{9x^{2}+27x}{9}=\frac{0}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$x^{2}+\frac{27}{9}x=\frac{0}{9}$$

Divide 27 by 9.

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

Divide 0 by 9.

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

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

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

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

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

Factor x^{2}+3x+\frac{9}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{3}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Simplify.

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

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

$$x=0 x=-3$$