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

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

Solve for x (complex solution)

$$x=1+\sqrt{2}i\approx 1+1.414213562i$$

$$x=-\sqrt{2}i+1\approx 1-1.414213562i$$

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}.

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

Add 2x^{2} to both sides.

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

Combine x^{2} and 2x^{2} to get 3x^{2}.

$$3x^{2}-6x+9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -6 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 3\times 9}}{2\times 3}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times 3\times 9}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-6\right)±\sqrt{36-12\times 9}}{2\times 3}$$

Multiply -12 times 9.

$$x=\frac{-\left(-6\right)±\sqrt{36-108}}{2\times 3}$$

Add 36 to -108.

$$x=\frac{-\left(-6\right)±\sqrt{-72}}{2\times 3}$$

Take the square root of -72.

$$x=\frac{-\left(-6\right)±6\sqrt{2}i}{2\times 3}$$

The opposite of -6 is 6.

$$x=\frac{6±6\sqrt{2}i}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{6±6\sqrt{2}i}{6}$$

Now solve the equation x=\frac{6±6\sqrt{2}i}{6} when ± is plus. Add 6 to 6i\sqrt{2}.

$$x=\frac{6+6\sqrt{2}i}{6}$$

Divide 6+6i\sqrt{2} by 6.

$$x=1+\sqrt{2}i$$

Now solve the equation x=\frac{6±6\sqrt{2}i}{6} when ± is minus. Subtract 6i\sqrt{2} from 6.

$$x=\frac{-6\sqrt{2}i+6}{6}$$

Divide 6-6i\sqrt{2} by 6.

$$x=-\sqrt{2}i+1$$

The equation is now solved.

$$x=1+\sqrt{2}i x=-\sqrt{2}i+1$$

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}.

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

Add 2x^{2} to both sides.

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

Combine x^{2} and 2x^{2} to get 3x^{2}.

$$3x^{2}-6x+9=0$$

Subtract 9 from both sides. Anything subtracted from zero gives its negation.

$$3x^{2}-6x=-9$$

Divide both sides by 3.

$$\frac{3x^{2}-6x}{3}=-\frac{9}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\left(-\frac{6}{3}\right)x=-\frac{9}{3}$$

Divide -6 by 3.

$$x^{2}-2x=-\frac{9}{3}$$

Divide -9 by 3.

$$x^{2}-2x=-3$$

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.

$$x^{2}-2x+1=-3+1$$

Add -3 to 1.

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-1=\sqrt{2}i x-1=-\sqrt{2}i$$

Add 1 to both sides of the equation.

$$x=1+\sqrt{2}i x=-\sqrt{2}i+1$$