Solve (x+1)(9x+9)=2

$$( x + 1 ) ( 9 x + 9 ) = 2$$

Solve for x

$$x=\frac{\sqrt{2}}{3}-1\approx -0.528595479$$

$$x=-\frac{\sqrt{2}}{3}-1\approx -1.471404521$$

Steps Using the Quadratic Formula

Use the distributive property to multiply x+1 by 9x+9 and combine like terms.

$$9x^{2}+18x+9=2$$

Subtract 2 from both sides.

$$9x^{2}+18x+9-2=0$$

Subtract 2 from 9 to get 7.

$$9x^{2}+18x+7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, 18 for b, and 7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-18±\sqrt{18^{2}-4\times 9\times 7}}{2\times 9}$$

Square 18.

$$x=\frac{-18±\sqrt{324-4\times 9\times 7}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-18±\sqrt{324-36\times 7}}{2\times 9}$$

Multiply -36 times 7.

$$x=\frac{-18±\sqrt{324-252}}{2\times 9}$$

Add 324 to -252.

$$x=\frac{-18±\sqrt{72}}{2\times 9}$$

Take the square root of 72.

$$x=\frac{-18±6\sqrt{2}}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{-18±6\sqrt{2}}{18}$$

Now solve the equation x=\frac{-18±6\sqrt{2}}{18} when ± is plus. Add -18 to 6\sqrt{2}.

$$x=\frac{6\sqrt{2}-18}{18}$$

Divide -18+6\sqrt{2} by 18.

$$x=\frac{\sqrt{2}}{3}-1$$

Now solve the equation x=\frac{-18±6\sqrt{2}}{18} when ± is minus. Subtract 6\sqrt{2} from -18.

$$x=\frac{-6\sqrt{2}-18}{18}$$

Divide -18-6\sqrt{2} by 18.

$$x=-\frac{\sqrt{2}}{3}-1$$

The equation is now solved.

$$x=\frac{\sqrt{2}}{3}-1 x=-\frac{\sqrt{2}}{3}-1$$

Steps for Completing the Square

Use the distributive property to multiply x+1 by 9x+9 and combine like terms.

$$9x^{2}+18x+9=2$$

Subtract 9 from both sides.

$$9x^{2}+18x=2-9$$

Subtract 9 from 2 to get -7.

$$9x^{2}+18x=-7$$

Divide both sides by 9.

$$\frac{9x^{2}+18x}{9}=-\frac{7}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$x^{2}+\frac{18}{9}x=-\frac{7}{9}$$

Divide 18 by 9.

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

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

$$x^{2}+2x+1^{2}=-\frac{7}{9}+1^{2}$$

Square 1.

$$x^{2}+2x+1=-\frac{7}{9}+1$$

Add -\frac{7}{9} to 1.

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

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

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

Take the square root of both sides of the equation.

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

Simplify.

$$x+1=\frac{\sqrt{2}}{3} x+1=-\frac{\sqrt{2}}{3}$$

Subtract 1 from both sides of the equation.

$$x=\frac{\sqrt{2}}{3}-1 x=-\frac{\sqrt{2}}{3}-1$$