Solve (x+1)^2=5+2x(x-1)=

$$( x + 1 ) ^ { 2 } = 5 + 2 x ( x - 1 ) =$$

Solve for x

$$x=2$$

Steps Using Factoring By Grouping

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

$$x^{2}+2x+1=5+2x\left(x-1\right)$$

Use the distributive property to multiply 2x by x-1.

$$x^{2}+2x+1=5+2x^{2}-2x$$

Subtract 5 from both sides.

$$x^{2}+2x+1-5=2x^{2}-2x$$

Subtract 5 from 1 to get -4.

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

Subtract 2x^{2} from both sides.

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

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

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

Add 2x to both sides.

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

Combine 2x and 2x to get 4x.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-4. To find a and b, set up a system to be solved.

$$a+b=4 ab=-\left(-4\right)=4$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 4.

$$1,4 2,2$$

Calculate the sum for each pair.

$$1+4=5 2+2=4$$

The solution is the pair that gives sum 4.

$$a=2 b=2$$

Rewrite -x^{2}+4x-4 as \left(-x^{2}+2x\right)+\left(2x-4\right).

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

Factor out -x in the first and 2 in the second group.

$$-x\left(x-2\right)+2\left(x-2\right)$$

Factor out common term x-2 by using distributive property.

$$\left(x-2\right)\left(-x+2\right)$$

To find equation solutions, solve x-2=0 and -x+2=0.

$$x=2 x=2$$

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}.

$$x^{2}+2x+1=5+2x\left(x-1\right)$$

Use the distributive property to multiply 2x by x-1.

$$x^{2}+2x+1=5+2x^{2}-2x$$

Subtract 5 from both sides.

$$x^{2}+2x+1-5=2x^{2}-2x$$

Subtract 5 from 1 to get -4.

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

Subtract 2x^{2} from both sides.

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

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

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

Add 2x to both sides.

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

Combine 2x and 2x to get 4x.

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

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

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

Square 4.

$$x=\frac{-4±\sqrt{16-4\left(-1\right)\left(-4\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-4±\sqrt{16+4\left(-4\right)}}{2\left(-1\right)}$$

Multiply 4 times -4.

$$x=\frac{-4±\sqrt{16-16}}{2\left(-1\right)}$$

Add 16 to -16.

$$x=\frac{-4±\sqrt{0}}{2\left(-1\right)}$$

Take the square root of 0.

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

Multiply 2 times -1.

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

Divide -4 by -2.

$$x=2$$

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}.

$$x^{2}+2x+1=5+2x\left(x-1\right)$$

Use the distributive property to multiply 2x by x-1.

$$x^{2}+2x+1=5+2x^{2}-2x$$

Subtract 2x^{2} from both sides.

$$x^{2}+2x+1-2x^{2}=5-2x$$

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

$$-x^{2}+2x+1=5-2x$$

Add 2x to both sides.

$$-x^{2}+2x+1+2x=5$$

Combine 2x and 2x to get 4x.

$$-x^{2}+4x+1=5$$

Subtract 1 from both sides.

$$-x^{2}+4x=5-1$$

Subtract 1 from 5 to get 4.

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

Divide both sides by -1.

$$\frac{-x^{2}+4x}{-1}=\frac{4}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{4}{-1}x=\frac{4}{-1}$$

Divide 4 by -1.

$$x^{2}-4x=\frac{4}{-1}$$

Divide 4 by -1.

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

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.

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

Square -2.

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

Add -4 to 4.

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

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}.

$$\left(x-2\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{0}$$

Simplify.

$$x-2=0 x-2=0$$

Add 2 to both sides of the equation.

$$x=2 x=2$$

The equation is now solved. Solutions are the same.

$$x=2$$