Solve (x)^2=(x+11)^2+(x-3)^2

$$( x ) ^ { 2 } = ( x + 11 ) ^ { 2 } + ( x - 3 ) ^ { 2 }$$

Solve for x (complex solution)

$$x=-\sqrt{66}i-8\approx -8-8.124038405i$$

$$x=-8+\sqrt{66}i\approx -8+8.124038405i$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+11\right)^{2}.

$$x^{2}=x^{2}+22x+121+\left(x-3\right)^{2}$$

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

$$x^{2}=x^{2}+22x+121+x^{2}-6x+9$$

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

$$x^{2}=2x^{2}+22x+121-6x+9$$

Combine 22x and -6x to get 16x.

$$x^{2}=2x^{2}+16x+121+9$$

Add 121 and 9 to get 130.

$$x^{2}=2x^{2}+16x+130$$

Subtract 2x^{2} from both sides.

$$x^{2}-2x^{2}=16x+130$$

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

$$-x^{2}=16x+130$$

Subtract 16x from both sides.

$$-x^{2}-16x=130$$

Subtract 130 from both sides.

$$-x^{2}-16x-130=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -16 for b, and -130 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\left(-1\right)\left(-130\right)}}{2\left(-1\right)}$$

Square -16.

$$x=\frac{-\left(-16\right)±\sqrt{256-4\left(-1\right)\left(-130\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-16\right)±\sqrt{256+4\left(-130\right)}}{2\left(-1\right)}$$

Multiply 4 times -130.

$$x=\frac{-\left(-16\right)±\sqrt{256-520}}{2\left(-1\right)}$$

Add 256 to -520.

$$x=\frac{-\left(-16\right)±\sqrt{-264}}{2\left(-1\right)}$$

Take the square root of -264.

$$x=\frac{-\left(-16\right)±2\sqrt{66}i}{2\left(-1\right)}$$

The opposite of -16 is 16.

$$x=\frac{16±2\sqrt{66}i}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{16±2\sqrt{66}i}{-2}$$

Now solve the equation x=\frac{16±2\sqrt{66}i}{-2} when ± is plus. Add 16 to 2i\sqrt{66}.

$$x=\frac{16+2\sqrt{66}i}{-2}$$

Divide 16+2i\sqrt{66} by -2.

$$x=-\sqrt{66}i-8$$

Now solve the equation x=\frac{16±2\sqrt{66}i}{-2} when ± is minus. Subtract 2i\sqrt{66} from 16.

$$x=\frac{-2\sqrt{66}i+16}{-2}$$

Divide 16-2i\sqrt{66} by -2.

$$x=-8+\sqrt{66}i$$

The equation is now solved.

$$x=-\sqrt{66}i-8 x=-8+\sqrt{66}i$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+11\right)^{2}.

$$x^{2}=x^{2}+22x+121+\left(x-3\right)^{2}$$

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

$$x^{2}=x^{2}+22x+121+x^{2}-6x+9$$

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

$$x^{2}=2x^{2}+22x+121-6x+9$$

Combine 22x and -6x to get 16x.

$$x^{2}=2x^{2}+16x+121+9$$

Add 121 and 9 to get 130.

$$x^{2}=2x^{2}+16x+130$$

Subtract 2x^{2} from both sides.

$$x^{2}-2x^{2}=16x+130$$

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

$$-x^{2}=16x+130$$

Subtract 16x from both sides.

$$-x^{2}-16x=130$$

Divide both sides by -1.

$$\frac{-x^{2}-16x}{-1}=\frac{130}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{16}{-1}\right)x=\frac{130}{-1}$$

Divide -16 by -1.

$$x^{2}+16x=\frac{130}{-1}$$

Divide 130 by -1.

$$x^{2}+16x=-130$$

Divide 16, the coefficient of the x term, by 2 to get 8. Then add the square of 8 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+16x+8^{2}=-130+8^{2}$$

Square 8.

$$x^{2}+16x+64=-130+64$$

Add -130 to 64.

$$x^{2}+16x+64=-66$$

Factor x^{2}+16x+64. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+8\right)^{2}=-66$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+8\right)^{2}}=\sqrt{-66}$$

Simplify.

$$x+8=\sqrt{66}i x+8=-\sqrt{66}i$$

Subtract 8 from both sides of the equation.

$$x=-8+\sqrt{66}i x=-\sqrt{66}i-8$$