Solve left(x+3right)^2+x^2=15^2

$${ \left(x+3 \right) }^{ 2 } + { x }^{ 2 } = { 15 }^{ 2 }$$

Solve for x

$$x=-12$$

$$x=9$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$x^{2}+6x+9+x^{2}=15^{2}$$

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

$$2x^{2}+6x+9=15^{2}$$

Calculate 15 to the power of 2 and get 225.

$$2x^{2}+6x+9=225$$

Subtract 225 from both sides.

$$2x^{2}+6x+9-225=0$$

Subtract 225 from 9 to get -216.

$$2x^{2}+6x-216=0$$

Divide both sides by 2.

$$x^{2}+3x-108=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-108. To find a and b, set up a system to be solved.

$$a+b=3 ab=1\left(-108\right)=-108$$

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 -108.

$$-1,108 -2,54 -3,36 -4,27 -6,18 -9,12$$

Calculate the sum for each pair.

$$-1+108=107 -2+54=52 -3+36=33 -4+27=23 -6+18=12 -9+12=3$$

The solution is the pair that gives sum 3.

$$a=-9 b=12$$

Rewrite x^{2}+3x-108 as \left(x^{2}-9x\right)+\left(12x-108\right).

$$\left(x^{2}-9x\right)+\left(12x-108\right)$$

Factor out x in the first and 12 in the second group.

$$x\left(x-9\right)+12\left(x-9\right)$$

Factor out common term x-9 by using distributive property.

$$\left(x-9\right)\left(x+12\right)$$

To find equation solutions, solve x-9=0 and x+12=0.

$$x=9 x=-12$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$x^{2}+6x+9+x^{2}=15^{2}$$

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

$$2x^{2}+6x+9=15^{2}$$

Calculate 15 to the power of 2 and get 225.

$$2x^{2}+6x+9=225$$

Subtract 225 from both sides.

$$2x^{2}+6x+9-225=0$$

Subtract 225 from 9 to get -216.

$$2x^{2}+6x-216=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 6 for b, and -216 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\times 2\left(-216\right)}}{2\times 2}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\times 2\left(-216\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-6±\sqrt{36-8\left(-216\right)}}{2\times 2}$$

Multiply -8 times -216.

$$x=\frac{-6±\sqrt{36+1728}}{2\times 2}$$

Add 36 to 1728.

$$x=\frac{-6±\sqrt{1764}}{2\times 2}$$

Take the square root of 1764.

$$x=\frac{-6±42}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-6±42}{4}$$

Now solve the equation x=\frac{-6±42}{4} when ± is plus. Add -6 to 42.

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

Divide 36 by 4.

$$x=9$$

Now solve the equation x=\frac{-6±42}{4} when ± is minus. Subtract 42 from -6.

$$x=-\frac{48}{4}$$

Divide -48 by 4.

$$x=-12$$

The equation is now solved.

$$x=9 x=-12$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$x^{2}+6x+9+x^{2}=15^{2}$$

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

$$2x^{2}+6x+9=15^{2}$$

Calculate 15 to the power of 2 and get 225.

$$2x^{2}+6x+9=225$$

Subtract 9 from both sides.

$$2x^{2}+6x=225-9$$

Subtract 9 from 225 to get 216.

$$2x^{2}+6x=216$$

Divide both sides by 2.

$$\frac{2x^{2}+6x}{2}=\frac{216}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{6}{2}x=\frac{216}{2}$$

Divide 6 by 2.

$$x^{2}+3x=\frac{216}{2}$$

Divide 216 by 2.

$$x^{2}+3x=108$$

Divide 3, the coefficient of the x term, by 2 to get \frac{3}{2}. Then add the square of \frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+3x+\left(\frac{3}{2}\right)^{2}=108+\left(\frac{3}{2}\right)^{2}$$

Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+3x+\frac{9}{4}=108+\frac{9}{4}$$

Add 108 to \frac{9}{4}.

$$x^{2}+3x+\frac{9}{4}=\frac{441}{4}$$

Factor x^{2}+3x+\frac{9}{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}.

$$\left(x+\frac{3}{2}\right)^{2}=\frac{441}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{\frac{441}{4}}$$

Simplify.

$$x+\frac{3}{2}=\frac{21}{2} x+\frac{3}{2}=-\frac{21}{2}$$

Subtract \frac{3}{2} from both sides of the equation.

$$x=9 x=-12$$