Solve x^2+left(x+2right)^2=130

$${ x }^{ 2 } + { \left(x+2 \right) }^{ 2 } =130$$

Solve for x

$$x=-9$$

$$x=7$$

Steps Using Factoring By Grouping

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

$$x^{2}+x^{2}+4x+4=130$$

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

$$2x^{2}+4x+4=130$$

Subtract 130 from both sides.

$$2x^{2}+4x+4-130=0$$

Subtract 130 from 4 to get -126.

$$2x^{2}+4x-126=0$$

Divide both sides by 2.

$$x^{2}+2x-63=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-63. To find a and b, set up a system to be solved.

$$a+b=2 ab=1\left(-63\right)=-63$$

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. List all such integer pairs that give product -63.

$$-1,63 -3,21 -7,9$$

Calculate the sum for each pair.

$$-1+63=62 -3+21=18 -7+9=2$$

The solution is the pair that gives sum 2.

$$a=-7 b=9$$

Rewrite x^{2}+2x-63 as \left(x^{2}-7x\right)+\left(9x-63\right).

$$\left(x^{2}-7x\right)+\left(9x-63\right)$$

Factor out x in the first and 9 in the second group.

$$x\left(x-7\right)+9\left(x-7\right)$$

Factor out common term x-7 by using distributive property.

$$\left(x-7\right)\left(x+9\right)$$

To find equation solutions, solve x-7=0 and x+9=0.

$$x=7 x=-9$$

Steps Using the Quadratic Formula

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

$$x^{2}+x^{2}+4x+4=130$$

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

$$2x^{2}+4x+4=130$$

Subtract 130 from both sides.

$$2x^{2}+4x+4-130=0$$

Subtract 130 from 4 to get -126.

$$2x^{2}+4x-126=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 4 for b, and -126 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-4±\sqrt{4^{2}-4\times 2\left(-126\right)}}{2\times 2}$$

Square 4.

$$x=\frac{-4±\sqrt{16-4\times 2\left(-126\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-4±\sqrt{16-8\left(-126\right)}}{2\times 2}$$

Multiply -8 times -126.

$$x=\frac{-4±\sqrt{16+1008}}{2\times 2}$$

Add 16 to 1008.

$$x=\frac{-4±\sqrt{1024}}{2\times 2}$$

Take the square root of 1024.

$$x=\frac{-4±32}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-4±32}{4}$$

Now solve the equation x=\frac{-4±32}{4} when ± is plus. Add -4 to 32.

$$x=\frac{28}{4}$$

Divide 28 by 4.

$$x=7$$

Now solve the equation x=\frac{-4±32}{4} when ± is minus. Subtract 32 from -4.

$$x=-\frac{36}{4}$$

Divide -36 by 4.

$$x=-9$$

The equation is now solved.

$$x=7 x=-9$$

Steps for Completing the Square

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

$$x^{2}+x^{2}+4x+4=130$$

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

$$2x^{2}+4x+4=130$$

Subtract 4 from both sides.

$$2x^{2}+4x=130-4$$

Subtract 4 from 130 to get 126.

$$2x^{2}+4x=126$$

Divide both sides by 2.

$$\frac{2x^{2}+4x}{2}=\frac{126}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{4}{2}x=\frac{126}{2}$$

Divide 4 by 2.

$$x^{2}+2x=\frac{126}{2}$$

Divide 126 by 2.

$$x^{2}+2x=63$$

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.

$$x^{2}+2x+1^{2}=63+1^{2}$$

Square 1.

$$x^{2}+2x+1=63+1$$

Add 63 to 1.

$$x^{2}+2x+1=64$$

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}.

$$\left(x+1\right)^{2}=64$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{64}$$

Simplify.

$$x+1=8 x+1=-8$$

Subtract 1 from both sides of the equation.

$$x=7 x=-9$$