Solve x^2+left(x+2right)^2=580

$${ x }^{ 2 } + { \left(x+2 \right) }^{ 2 } =580$$

Solve for x

$$x=-18$$

$$x=16$$

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}.

$$x^{2}+x^{2}+4x+4=580$$

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

$$2x^{2}+4x+4=580$$

Subtract 580 from both sides.

$$2x^{2}+4x+4-580=0$$

Subtract 580 from 4 to get -576.

$$2x^{2}+4x-576=0$$

Divide both sides by 2.

$$x^{2}+2x-288=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-288. To find a and b, set up a system to be solved.

$$a+b=2 ab=1\left(-288\right)=-288$$

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 -288.

$$-1,288 -2,144 -3,96 -4,72 -6,48 -8,36 -9,32 -12,24 -16,18$$

Calculate the sum for each pair.

$$-1+288=287 -2+144=142 -3+96=93 -4+72=68 -6+48=42 -8+36=28 -9+32=23 -12+24=12 -16+18=2$$

The solution is the pair that gives sum 2.

$$a=-16 b=18$$

Rewrite x^{2}+2x-288 as \left(x^{2}-16x\right)+\left(18x-288\right).

$$\left(x^{2}-16x\right)+\left(18x-288\right)$$

Factor out x in the first and 18 in the second group.

$$x\left(x-16\right)+18\left(x-16\right)$$

Factor out common term x-16 by using distributive property.

$$\left(x-16\right)\left(x+18\right)$$

To find equation solutions, solve x-16=0 and x+18=0.

$$x=16 x=-18$$

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}.

$$x^{2}+x^{2}+4x+4=580$$

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

$$2x^{2}+4x+4=580$$

Subtract 580 from both sides.

$$2x^{2}+4x+4-580=0$$

Subtract 580 from 4 to get -576.

$$2x^{2}+4x-576=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 4 for b, and -576 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-4±\sqrt{4^{2}-4\times 2\left(-576\right)}}{2\times 2}$$

Square 4.

$$x=\frac{-4±\sqrt{16-4\times 2\left(-576\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-4±\sqrt{16-8\left(-576\right)}}{2\times 2}$$

Multiply -8 times -576.

$$x=\frac{-4±\sqrt{16+4608}}{2\times 2}$$

Add 16 to 4608.

$$x=\frac{-4±\sqrt{4624}}{2\times 2}$$

Take the square root of 4624.

$$x=\frac{-4±68}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-4±68}{4}$$

Now solve the equation x=\frac{-4±68}{4} when ± is plus. Add -4 to 68.

$$x=\frac{64}{4}$$

Divide 64 by 4.

$$x=16$$

Now solve the equation x=\frac{-4±68}{4} when ± is minus. Subtract 68 from -4.

$$x=-\frac{72}{4}$$

Divide -72 by 4.

$$x=-18$$

The equation is now solved.

$$x=16 x=-18$$

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}.

$$x^{2}+x^{2}+4x+4=580$$

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

$$2x^{2}+4x+4=580$$

Subtract 4 from both sides.

$$2x^{2}+4x=580-4$$

Subtract 4 from 580 to get 576.

$$2x^{2}+4x=576$$

Divide both sides by 2.

$$\frac{2x^{2}+4x}{2}=\frac{576}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{4}{2}x=\frac{576}{2}$$

Divide 4 by 2.

$$x^{2}+2x=\frac{576}{2}$$

Divide 576 by 2.

$$x^{2}+2x=288$$

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.

$$x^{2}+2x+1^{2}=288+1^{2}$$

Square 1.

$$x^{2}+2x+1=288+1$$

Add 288 to 1.

$$x^{2}+2x+1=289$$

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}.

$$\left(x+1\right)^{2}=289$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{289}$$

Simplify.

$$x+1=17 x+1=-17$$

Subtract 1 from both sides of the equation.

$$x=16 x=-18$$