Solve x^2+(20-x)^2=17

$$x ^ { 2 } + ( 20 - x ) ^ { 2 } = 17$$

Solve for x (complex solution)

$$x=\frac{\sqrt{366}i}{2}+10\approx 10+9.565563235i$$

$$x=-\frac{\sqrt{366}i}{2}+10\approx 10-9.565563235i$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(20-x\right)^{2}.

$$x^{2}+400-40x+x^{2}=17$$

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

$$2x^{2}+400-40x=17$$

Subtract 17 from both sides.

$$2x^{2}+400-40x-17=0$$

Subtract 17 from 400 to get 383.

$$2x^{2}+383-40x=0$$

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.

$$2x^{2}-40x+383=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -40 for b, and 383 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-40\right)±\sqrt{\left(-40\right)^{2}-4\times 2\times 383}}{2\times 2}$$

Square -40.

$$x=\frac{-\left(-40\right)±\sqrt{1600-4\times 2\times 383}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-40\right)±\sqrt{1600-8\times 383}}{2\times 2}$$

Multiply -8 times 383.

$$x=\frac{-\left(-40\right)±\sqrt{1600-3064}}{2\times 2}$$

Add 1600 to -3064.

$$x=\frac{-\left(-40\right)±\sqrt{-1464}}{2\times 2}$$

Take the square root of -1464.

$$x=\frac{-\left(-40\right)±2\sqrt{366}i}{2\times 2}$$

The opposite of -40 is 40.

$$x=\frac{40±2\sqrt{366}i}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{40±2\sqrt{366}i}{4}$$

Now solve the equation x=\frac{40±2\sqrt{366}i}{4} when ± is plus. Add 40 to 2i\sqrt{366}.

$$x=\frac{40+2\sqrt{366}i}{4}$$

Divide 40+2i\sqrt{366} by 4.

$$x=\frac{\sqrt{366}i}{2}+10$$

Now solve the equation x=\frac{40±2\sqrt{366}i}{4} when ± is minus. Subtract 2i\sqrt{366} from 40.

$$x=\frac{-2\sqrt{366}i+40}{4}$$

Divide 40-2i\sqrt{366} by 4.

$$x=-\frac{\sqrt{366}i}{2}+10$$

The equation is now solved.

$$x=\frac{\sqrt{366}i}{2}+10 x=-\frac{\sqrt{366}i}{2}+10$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(20-x\right)^{2}.

$$x^{2}+400-40x+x^{2}=17$$

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

$$2x^{2}+400-40x=17$$

Subtract 400 from both sides.

$$2x^{2}-40x=17-400$$

Subtract 400 from 17 to get -383.

$$2x^{2}-40x=-383$$

Divide both sides by 2.

$$\frac{2x^{2}-40x}{2}=-\frac{383}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{40}{2}\right)x=-\frac{383}{2}$$

Divide -40 by 2.

$$x^{2}-20x=-\frac{383}{2}$$

Divide -20, the coefficient of the x term, by 2 to get -10. Then add the square of -10 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-20x+\left(-10\right)^{2}=-\frac{383}{2}+\left(-10\right)^{2}$$

Square -10.

$$x^{2}-20x+100=-\frac{383}{2}+100$$

Add -\frac{383}{2} to 100.

$$x^{2}-20x+100=-\frac{183}{2}$$

Factor x^{2}-20x+100. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-10\right)^{2}=-\frac{183}{2}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-10\right)^{2}}=\sqrt{-\frac{183}{2}}$$

Simplify.

$$x-10=\frac{\sqrt{366}i}{2} x-10=-\frac{\sqrt{366}i}{2}$$

Add 10 to both sides of the equation.

$$x=\frac{\sqrt{366}i}{2}+10 x=-\frac{\sqrt{366}i}{2}+10$$