Solve x^2-18x+209=2x+11

$$x ^ { 2 } - 18 x + 209 = 2 x + 11$$

Solve for x (complex solution)

$$x=10+7\sqrt{2}i\approx 10+9.899494937i$$

$$x=-7\sqrt{2}i+10\approx 10-9.899494937i$$

Steps Using the Quadratic Formula

Subtract 2x from both sides.

$$x^{2}-18x+209-2x=11$$

Combine -18x and -2x to get -20x.

$$x^{2}-20x+209=11$$

Subtract 11 from both sides.

$$x^{2}-20x+209-11=0$$

Subtract 11 from 209 to get 198.

$$x^{2}-20x+198=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -20 for b, and 198 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 198}}{2}$$

Square -20.

$$x=\frac{-\left(-20\right)±\sqrt{400-4\times 198}}{2}$$

Multiply -4 times 198.

$$x=\frac{-\left(-20\right)±\sqrt{400-792}}{2}$$

Add 400 to -792.

$$x=\frac{-\left(-20\right)±\sqrt{-392}}{2}$$

Take the square root of -392.

$$x=\frac{-\left(-20\right)±14\sqrt{2}i}{2}$$

The opposite of -20 is 20.

$$x=\frac{20±14\sqrt{2}i}{2}$$

Now solve the equation x=\frac{20±14\sqrt{2}i}{2} when ± is plus. Add 20 to 14i\sqrt{2}.

$$x=\frac{20+14\sqrt{2}i}{2}$$

Divide 20+14i\sqrt{2} by 2.

$$x=10+7\sqrt{2}i$$

Now solve the equation x=\frac{20±14\sqrt{2}i}{2} when ± is minus. Subtract 14i\sqrt{2} from 20.

$$x=\frac{-14\sqrt{2}i+20}{2}$$

Divide 20-14i\sqrt{2} by 2.

$$x=-7\sqrt{2}i+10$$

The equation is now solved.

$$x=10+7\sqrt{2}i x=-7\sqrt{2}i+10$$

Steps for Completing the Square

Subtract 2x from both sides.

$$x^{2}-18x+209-2x=11$$

Combine -18x and -2x to get -20x.

$$x^{2}-20x+209=11$$

Subtract 209 from both sides.

$$x^{2}-20x=11-209$$

Subtract 209 from 11 to get -198.

$$x^{2}-20x=-198$$

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.

$$x^{2}-20x+\left(-10\right)^{2}=-198+\left(-10\right)^{2}$$

Square -10.

$$x^{2}-20x+100=-198+100$$

Add -198 to 100.

$$x^{2}-20x+100=-98$$

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}.

$$\left(x-10\right)^{2}=-98$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-10\right)^{2}}=\sqrt{-98}$$

Simplify.

$$x-10=7\sqrt{2}i x-10=-7\sqrt{2}i$$

Add 10 to both sides of the equation.

$$x=10+7\sqrt{2}i x=-7\sqrt{2}i+10$$