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

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

Solve for x

$$x=2\sqrt{6}-1\approx 3.898979486$$

$$x=-2\sqrt{6}-1\approx -5.898979486$$

Steps Using the Quadratic Formula

Add 4 and 8 to get 12.

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

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

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

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

$$12=\frac{1}{2}x^{2}+x+\frac{1}{2}$$

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

$$\frac{1}{2}x^{2}+x+\frac{1}{2}=12$$

Subtract 12 from both sides.

$$\frac{1}{2}x^{2}+x+\frac{1}{2}-12=0$$

Subtract 12 from \frac{1}{2} to get -\frac{23}{2}.

$$\frac{1}{2}x^{2}+x-\frac{23}{2}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{2} for a, 1 for b, and -\frac{23}{2} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}-4\times \frac{1}{2}\left(-\frac{23}{2}\right)}}{2\times \frac{1}{2}}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\times \frac{1}{2}\left(-\frac{23}{2}\right)}}{2\times \frac{1}{2}}$$

Multiply -4 times \frac{1}{2}.

$$x=\frac{-1±\sqrt{1-2\left(-\frac{23}{2}\right)}}{2\times \frac{1}{2}}$$

Multiply -2 times -\frac{23}{2}.

$$x=\frac{-1±\sqrt{1+23}}{2\times \frac{1}{2}}$$

Add 1 to 23.

$$x=\frac{-1±\sqrt{24}}{2\times \frac{1}{2}}$$

Take the square root of 24.

$$x=\frac{-1±2\sqrt{6}}{2\times \frac{1}{2}}$$

Multiply 2 times \frac{1}{2}.

$$x=\frac{-1±2\sqrt{6}}{1}$$

Now solve the equation x=\frac{-1±2\sqrt{6}}{1} when ± is plus. Add -1 to 2\sqrt{6}.

$$x=\frac{2\sqrt{6}-1}{1}$$

Divide -1+2\sqrt{6} by 1.

$$x=2\sqrt{6}-1$$

Now solve the equation x=\frac{-1±2\sqrt{6}}{1} when ± is minus. Subtract 2\sqrt{6} from -1.

$$x=\frac{-2\sqrt{6}-1}{1}$$

Divide -1-2\sqrt{6} by 1.

$$x=-2\sqrt{6}-1$$

The equation is now solved.

$$x=2\sqrt{6}-1 x=-2\sqrt{6}-1$$

Steps for Completing the Square

Add 4 and 8 to get 12.

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

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

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

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

$$12=\frac{1}{2}x^{2}+x+\frac{1}{2}$$

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

$$\frac{1}{2}x^{2}+x+\frac{1}{2}=12$$

Subtract \frac{1}{2} from both sides.

$$\frac{1}{2}x^{2}+x=12-\frac{1}{2}$$

Subtract \frac{1}{2} from 12 to get \frac{23}{2}.

$$\frac{1}{2}x^{2}+x=\frac{23}{2}$$

Multiply both sides by 2.

$$\frac{\frac{1}{2}x^{2}+x}{\frac{1}{2}}=\frac{\frac{23}{2}}{\frac{1}{2}}$$

Dividing by \frac{1}{2} undoes the multiplication by \frac{1}{2}.

$$x^{2}+\frac{1}{\frac{1}{2}}x=\frac{\frac{23}{2}}{\frac{1}{2}}$$

Divide 1 by \frac{1}{2} by multiplying 1 by the reciprocal of \frac{1}{2}.

$$x^{2}+2x=\frac{\frac{23}{2}}{\frac{1}{2}}$$

Divide \frac{23}{2} by \frac{1}{2} by multiplying \frac{23}{2} by the reciprocal of \frac{1}{2}.

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

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.

$$x^{2}+2x+1^{2}=23+1^{2}$$

Square 1.

$$x^{2}+2x+1=23+1$$

Add 23 to 1.

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

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}.

$$\left(x+1\right)^{2}=24$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{24}$$

Simplify.

$$x+1=2\sqrt{6} x+1=-2\sqrt{6}$$

Subtract 1 from both sides of the equation.

$$x=2\sqrt{6}-1 x=-2\sqrt{6}-1$$