Solve 0=9left(x+1right)^2-8

$$0=9 { \left(x+1 \right) }^{ 2 } -8$$

Solve for x

$$x=\frac{2\sqrt{2}}{3}-1\approx -0.057190958$$

$$x=-\frac{2\sqrt{2}}{3}-1\approx -1.942809042$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$0=9\left(x^{2}+2x+1\right)-8$$

Use the distributive property to multiply 9 by x^{2}+2x+1.

$$0=9x^{2}+18x+9-8$$

Subtract 8 from 9 to get 1.

$$0=9x^{2}+18x+1$$

Swap sides so that all variable terms are on the left hand side.

$$9x^{2}+18x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, 18 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square 18.

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

Multiply -4 times 9.

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

Add 324 to -36.

$$x=\frac{-18±\sqrt{288}}{2\times 9}$$

Take the square root of 288.

$$x=\frac{-18±12\sqrt{2}}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{-18±12\sqrt{2}}{18}$$

Now solve the equation x=\frac{-18±12\sqrt{2}}{18} when ± is plus. Add -18 to 12\sqrt{2}.

$$x=\frac{12\sqrt{2}-18}{18}$$

Divide -18+12\sqrt{2} by 18.

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

Now solve the equation x=\frac{-18±12\sqrt{2}}{18} when ± is minus. Subtract 12\sqrt{2} from -18.

$$x=\frac{-12\sqrt{2}-18}{18}$$

Divide -18-12\sqrt{2} by 18.

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

The equation is now solved.

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

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$0=9\left(x^{2}+2x+1\right)-8$$

Use the distributive property to multiply 9 by x^{2}+2x+1.

$$0=9x^{2}+18x+9-8$$

Subtract 8 from 9 to get 1.

$$0=9x^{2}+18x+1$$

Swap sides so that all variable terms are on the left hand side.

$$9x^{2}+18x+1=0$$

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

$$9x^{2}+18x=-1$$

Divide both sides by 9.

$$\frac{9x^{2}+18x}{9}=-\frac{1}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$x^{2}+\frac{18}{9}x=-\frac{1}{9}$$

Divide 18 by 9.

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

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.

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

Square 1.

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

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

$$x^{2}+2x+1=\frac{8}{9}$$

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}.

$$\left(x+1\right)^{2}=\frac{8}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{\frac{8}{9}}$$

Simplify.

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

Subtract 1 from both sides of the equation.

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