Solve (x+1)^2+2(-8-4)=4

$$( x + 1 ) ^ { 2 } + 2 ( - 8 - 4 ) = 4$$

Solve for x

$$x=2\sqrt{7}-1\approx 4.291502622$$

$$x=-2\sqrt{7}-1\approx -6.291502622$$

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}.

$$x^{2}+2x+1+2\left(-8-4\right)=4$$

Subtract 4 from -8 to get -12.

$$x^{2}+2x+1+2\left(-12\right)=4$$

Multiply 2 and -12 to get -24.

$$x^{2}+2x+1-24=4$$

Subtract 24 from 1 to get -23.

$$x^{2}+2x-23=4$$

Subtract 4 from both sides.

$$x^{2}+2x-23-4=0$$

Subtract 4 from -23 to get -27.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 2 for b, and -27 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2±\sqrt{2^{2}-4\left(-27\right)}}{2}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\left(-27\right)}}{2}$$

Multiply -4 times -27.

$$x=\frac{-2±\sqrt{4+108}}{2}$$

Add 4 to 108.

$$x=\frac{-2±\sqrt{112}}{2}$$

Take the square root of 112.

$$x=\frac{-2±4\sqrt{7}}{2}$$

Now solve the equation x=\frac{-2±4\sqrt{7}}{2} when ± is plus. Add -2 to 4\sqrt{7}.

$$x=\frac{4\sqrt{7}-2}{2}$$

Divide -2+4\sqrt{7} by 2.

$$x=2\sqrt{7}-1$$

Now solve the equation x=\frac{-2±4\sqrt{7}}{2} when ± is minus. Subtract 4\sqrt{7} from -2.

$$x=\frac{-4\sqrt{7}-2}{2}$$

Divide -2-4\sqrt{7} by 2.

$$x=-2\sqrt{7}-1$$

The equation is now solved.

$$x=2\sqrt{7}-1 x=-2\sqrt{7}-1$$

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}.

$$x^{2}+2x+1+2\left(-8-4\right)=4$$

Subtract 4 from -8 to get -12.

$$x^{2}+2x+1+2\left(-12\right)=4$$

Multiply 2 and -12 to get -24.

$$x^{2}+2x+1-24=4$$

Subtract 24 from 1 to get -23.

$$x^{2}+2x-23=4$$

Add 23 to both sides.

$$x^{2}+2x=4+23$$

Add 4 and 23 to get 27.

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

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.

$$x^{2}+2x+1^{2}=27+1^{2}$$

Square 1.

$$x^{2}+2x+1=27+1$$

Add 27 to 1.

$$x^{2}+2x+1=28$$

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}.

$$\left(x+1\right)^{2}=28$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{28}$$

Simplify.

$$x+1=2\sqrt{7} x+1=-2\sqrt{7}$$

Subtract 1 from both sides of the equation.

$$x=2\sqrt{7}-1 x=-2\sqrt{7}-1$$