Solve 3(x+2)^2-2=10

$$3 ( x + 2 ) ^ { 2 } - 2 = 10$$

Solve for x

$$x=0$$

$$x=-4$$

Steps Using Factoring

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

$$3\left(x^{2}+4x+4\right)-2=10$$

Use the distributive property to multiply 3 by x^{2}+4x+4.

$$3x^{2}+12x+12-2=10$$

Subtract 2 from 12 to get 10.

$$3x^{2}+12x+10=10$$

Subtract 10 from both sides.

$$3x^{2}+12x+10-10=0$$

Subtract 10 from 10 to get 0.

$$3x^{2}+12x=0$$

Factor out x.

$$x\left(3x+12\right)=0$$

To find equation solutions, solve x=0 and 3x+12=0.

$$x=0 x=-4$$

Steps Using the Quadratic Formula

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

$$3\left(x^{2}+4x+4\right)-2=10$$

Use the distributive property to multiply 3 by x^{2}+4x+4.

$$3x^{2}+12x+12-2=10$$

Subtract 2 from 12 to get 10.

$$3x^{2}+12x+10=10$$

Subtract 10 from both sides.

$$3x^{2}+12x+10-10=0$$

Subtract 10 from 10 to get 0.

$$3x^{2}+12x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 12 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-12±\sqrt{12^{2}}}{2\times 3}$$

Take the square root of 12^{2}.

$$x=\frac{-12±12}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-12±12}{6}$$

Now solve the equation x=\frac{-12±12}{6} when ± is plus. Add -12 to 12.

$$x=\frac{0}{6}$$

Divide 0 by 6.

$$x=0$$

Now solve the equation x=\frac{-12±12}{6} when ± is minus. Subtract 12 from -12.

$$x=-\frac{24}{6}$$

Divide -24 by 6.

$$x=-4$$

The equation is now solved.

$$x=0 x=-4$$

Steps for Completing the Square

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

$$3\left(x^{2}+4x+4\right)-2=10$$

Use the distributive property to multiply 3 by x^{2}+4x+4.

$$3x^{2}+12x+12-2=10$$

Subtract 2 from 12 to get 10.

$$3x^{2}+12x+10=10$$

Subtract 10 from both sides.

$$3x^{2}+12x=10-10$$

Subtract 10 from 10 to get 0.

$$3x^{2}+12x=0$$

Divide both sides by 3.

$$\frac{3x^{2}+12x}{3}=\frac{0}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{12}{3}x=\frac{0}{3}$$

Divide 12 by 3.

$$x^{2}+4x=\frac{0}{3}$$

Divide 0 by 3.

$$x^{2}+4x=0$$

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

$$x^{2}+4x+2^{2}=2^{2}$$

Square 2.

$$x^{2}+4x+4=4$$

Factor x^{2}+4x+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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x+2=2 x+2=-2$$

Subtract 2 from both sides of the equation.

$$x=0 x=-4$$