Solve +(x+1)^2=(2x-2)(x+3)

$$+ ( x + 1 ) ^ { 2 } = ( 2 x - 2 ) ( x + 3 )$$

Solve for x

$$x=2\sqrt{2}-1\approx 1.828427125$$

$$x=-2\sqrt{2}-1\approx -3.828427125$$

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}.

$$x^{2}+2x+1=\left(2x-2\right)\left(x+3\right)$$

Use the distributive property to multiply 2x-2 by x+3 and combine like terms.

$$x^{2}+2x+1=2x^{2}+4x-6$$

Subtract 2x^{2} from both sides.

$$x^{2}+2x+1-2x^{2}=4x-6$$

Combine x^{2} and -2x^{2} to get -x^{2}.

$$-x^{2}+2x+1=4x-6$$

Subtract 4x from both sides.

$$-x^{2}+2x+1-4x=-6$$

Combine 2x and -4x to get -2x.

$$-x^{2}-2x+1=-6$$

Add 6 to both sides.

$$-x^{2}-2x+1+6=0$$

Add 1 and 6 to get 7.

$$-x^{2}-2x+7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -2 for b, and 7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\left(-1\right)\times 7}}{2\left(-1\right)}$$

Square -2.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\left(-1\right)\times 7}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-2\right)±\sqrt{4+4\times 7}}{2\left(-1\right)}$$

Multiply 4 times 7.

$$x=\frac{-\left(-2\right)±\sqrt{4+28}}{2\left(-1\right)}$$

Add 4 to 28.

$$x=\frac{-\left(-2\right)±\sqrt{32}}{2\left(-1\right)}$$

Take the square root of 32.

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

The opposite of -2 is 2.

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

Multiply 2 times -1.

$$x=\frac{2±4\sqrt{2}}{-2}$$

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

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

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

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

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

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

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

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

The equation is now solved.

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

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}.

$$x^{2}+2x+1=\left(2x-2\right)\left(x+3\right)$$

Use the distributive property to multiply 2x-2 by x+3 and combine like terms.

$$x^{2}+2x+1=2x^{2}+4x-6$$

Subtract 2x^{2} from both sides.

$$x^{2}+2x+1-2x^{2}=4x-6$$

Combine x^{2} and -2x^{2} to get -x^{2}.

$$-x^{2}+2x+1=4x-6$$

Subtract 4x from both sides.

$$-x^{2}+2x+1-4x=-6$$

Combine 2x and -4x to get -2x.

$$-x^{2}-2x+1=-6$$

Subtract 1 from both sides.

$$-x^{2}-2x=-6-1$$

Subtract 1 from -6 to get -7.

$$-x^{2}-2x=-7$$

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{2}{-1}\right)x=-\frac{7}{-1}$$

Divide -2 by -1.

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

Divide -7 by -1.

$$x^{2}+2x=7$$

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.

$$x^{2}+2x+1^{2}=7+1^{2}$$

Square 1.

$$x^{2}+2x+1=7+1$$

Add 7 to 1.

$$x^{2}+2x+1=8$$

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Subtract 1 from both sides of the equation.

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