Solve 4+(2-x)^2+36+x^2=20

$$4 + ( 2 - x ) ^ { 2 } + 36 + x ^ { 2 } = 20$$

Solve for x (complex solution)

$$x=1+\sqrt{11}i\approx 1+3.31662479i$$

$$x=-\sqrt{11}i+1\approx 1-3.31662479i$$

Steps Using the Quadratic Formula

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

$$4+4-4x+x^{2}+36+x^{2}=20$$

Add 4 and 4 to get 8.

$$8-4x+x^{2}+36+x^{2}=20$$

Add 8 and 36 to get 44.

$$44-4x+x^{2}+x^{2}=20$$

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

$$44-4x+2x^{2}=20$$

Subtract 20 from both sides.

$$44-4x+2x^{2}-20=0$$

Subtract 20 from 44 to get 24.

$$24-4x+2x^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$2x^{2}-4x+24=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -4 for b, and 24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 2\times 24}}{2\times 2}$$

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times 2\times 24}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-4\right)±\sqrt{16-8\times 24}}{2\times 2}$$

Multiply -8 times 24.

$$x=\frac{-\left(-4\right)±\sqrt{16-192}}{2\times 2}$$

Add 16 to -192.

$$x=\frac{-\left(-4\right)±\sqrt{-176}}{2\times 2}$$

Take the square root of -176.

$$x=\frac{-\left(-4\right)±4\sqrt{11}i}{2\times 2}$$

The opposite of -4 is 4.

$$x=\frac{4±4\sqrt{11}i}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{4±4\sqrt{11}i}{4}$$

Now solve the equation x=\frac{4±4\sqrt{11}i}{4} when ± is plus. Add 4 to 4i\sqrt{11}.

$$x=\frac{4+4\sqrt{11}i}{4}$$

Divide 4+4i\sqrt{11} by 4.

$$x=1+\sqrt{11}i$$

Now solve the equation x=\frac{4±4\sqrt{11}i}{4} when ± is minus. Subtract 4i\sqrt{11} from 4.

$$x=\frac{-4\sqrt{11}i+4}{4}$$

Divide 4-4i\sqrt{11} by 4.

$$x=-\sqrt{11}i+1$$

The equation is now solved.

$$x=1+\sqrt{11}i x=-\sqrt{11}i+1$$

Steps for Completing the Square

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

$$4+4-4x+x^{2}+36+x^{2}=20$$

Add 4 and 4 to get 8.

$$8-4x+x^{2}+36+x^{2}=20$$

Add 8 and 36 to get 44.

$$44-4x+x^{2}+x^{2}=20$$

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

$$44-4x+2x^{2}=20$$

Subtract 44 from both sides.

$$-4x+2x^{2}=20-44$$

Subtract 44 from 20 to get -24.

$$-4x+2x^{2}=-24$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$2x^{2}-4x=-24$$

Divide both sides by 2.

$$\frac{2x^{2}-4x}{2}=-\frac{24}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{4}{2}\right)x=-\frac{24}{2}$$

Divide -4 by 2.

$$x^{2}-2x=-\frac{24}{2}$$

Divide -24 by 2.

$$x^{2}-2x=-12$$

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.

$$x^{2}-2x+1=-12+1$$

Add -12 to 1.

$$x^{2}-2x+1=-11$$

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}.

$$\left(x-1\right)^{2}=-11$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-1\right)^{2}}=\sqrt{-11}$$

Simplify.

$$x-1=\sqrt{11}i x-1=-\sqrt{11}i$$

Add 1 to both sides of the equation.

$$x=1+\sqrt{11}i x=-\sqrt{11}i+1$$