Solve 2(x+1)^2-4=x(x+3)

$$2 ( x + 1 ) ^ { 2 } - 4 = x ( x + 3 )$$

Solve for x

$$x=-2$$

$$x=1$$

Steps Using Factoring

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

$$2\left(x^{2}+2x+1\right)-4=x\left(x+3\right)$$

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

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

Subtract 4 from 2 to get -2.

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

Use the distributive property to multiply x by x+3.

$$2x^{2}+4x-2=x^{2}+3x$$

Subtract x^{2} from both sides.

$$2x^{2}+4x-2-x^{2}=3x$$

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

$$x^{2}+4x-2=3x$$

Subtract 3x from both sides.

$$x^{2}+4x-2-3x=0$$

Combine 4x and -3x to get x.

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

To solve the equation, factor x^{2}+x-2 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=1 ab=-2$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.

$$a=-1 b=2$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-1\right)\left(x+2\right)$$

To find equation solutions, solve x-1=0 and x+2=0.

$$x=1 x=-2$$

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}.

$$2\left(x^{2}+2x+1\right)-4=x\left(x+3\right)$$

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

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

Subtract 4 from 2 to get -2.

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

Use the distributive property to multiply x by x+3.

$$2x^{2}+4x-2=x^{2}+3x$$

Subtract x^{2} from both sides.

$$2x^{2}+4x-2-x^{2}=3x$$

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

$$x^{2}+4x-2=3x$$

Subtract 3x from both sides.

$$x^{2}+4x-2-3x=0$$

Combine 4x and -3x to get x.

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

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

$$a+b=1 ab=1\left(-2\right)=-2$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.

$$a=-1 b=2$$

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

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

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

$$x\left(x-1\right)+2\left(x-1\right)$$

Factor out common term x-1 by using distributive property.

$$\left(x-1\right)\left(x+2\right)$$

To find equation solutions, solve x-1=0 and x+2=0.

$$x=1 x=-2$$

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}.

$$2\left(x^{2}+2x+1\right)-4=x\left(x+3\right)$$

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

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

Subtract 4 from 2 to get -2.

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

Use the distributive property to multiply x by x+3.

$$2x^{2}+4x-2=x^{2}+3x$$

Subtract x^{2} from both sides.

$$2x^{2}+4x-2-x^{2}=3x$$

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

$$x^{2}+4x-2=3x$$

Subtract 3x from both sides.

$$x^{2}+4x-2-3x=0$$

Combine 4x and -3x to get x.

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

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

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

Square 1.

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

Multiply -4 times -2.

$$x=\frac{-1±\sqrt{1+8}}{2}$$

Add 1 to 8.

$$x=\frac{-1±\sqrt{9}}{2}$$

Take the square root of 9.

$$x=\frac{-1±3}{2}$$

Now solve the equation x=\frac{-1±3}{2} when ± is plus. Add -1 to 3.

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

Divide 2 by 2.

$$x=1$$

Now solve the equation x=\frac{-1±3}{2} when ± is minus. Subtract 3 from -1.

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

Divide -4 by 2.

$$x=-2$$

The equation is now solved.

$$x=1 x=-2$$

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}.

$$2\left(x^{2}+2x+1\right)-4=x\left(x+3\right)$$

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

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

Subtract 4 from 2 to get -2.

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

Use the distributive property to multiply x by x+3.

$$2x^{2}+4x-2=x^{2}+3x$$

Subtract x^{2} from both sides.

$$2x^{2}+4x-2-x^{2}=3x$$

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

$$x^{2}+4x-2=3x$$

Subtract 3x from both sides.

$$x^{2}+4x-2-3x=0$$

Combine 4x and -3x to get x.

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

Add 2 to both sides. Anything plus zero gives itself.

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

Divide 1, the coefficient of the x term, by 2 to get \frac{1}{2}. Then add the square of \frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

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

Add 2 to \frac{1}{4}.

$$x^{2}+x+\frac{1}{4}=\frac{9}{4}$$

Factor x^{2}+x+\frac{1}{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}.

$$\left(x+\frac{1}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Simplify.

$$x+\frac{1}{2}=\frac{3}{2} x+\frac{1}{2}=-\frac{3}{2}$$

Subtract \frac{1}{2} from both sides of the equation.

$$x=1 x=-2$$