Solve (0.2-x)^2=2x^2

$$( 0.2 - x ) ^ { 2 } = 2 x ^ { 2 }$$

Solve for x

$$x=\frac{\sqrt{2}-1}{5}\approx 0.082842712$$

$$x=\frac{-\sqrt{2}-1}{5}\approx -0.482842712$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(0.2-x\right)^{2}.

$$0.04-0.4x+x^{2}=2x^{2}$$

Subtract 2x^{2} from both sides.

$$0.04-0.4x+x^{2}-2x^{2}=0$$

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

$$0.04-0.4x-x^{2}=0$$

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.

$$-x^{2}-0.4x+0.04=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -0.4 for b, and 0.04 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-0.4\right)±\sqrt{\left(-0.4\right)^{2}-4\left(-1\right)\times 0.04}}{2\left(-1\right)}$$

Square -0.4 by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\left(-0.4\right)±\sqrt{0.16-4\left(-1\right)\times 0.04}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-0.4\right)±\sqrt{0.16+4\times 0.04}}{2\left(-1\right)}$$

Multiply 4 times 0.04.

$$x=\frac{-\left(-0.4\right)±\sqrt{\frac{4+4}{25}}}{2\left(-1\right)}$$

Add 0.16 to 0.16 by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-0.4\right)±\sqrt{0.32}}{2\left(-1\right)}$$

Take the square root of 0.32.

$$x=\frac{-\left(-0.4\right)±\frac{2\sqrt{2}}{5}}{2\left(-1\right)}$$

The opposite of -0.4 is 0.4.

$$x=\frac{0.4±\frac{2\sqrt{2}}{5}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{0.4±\frac{2\sqrt{2}}{5}}{-2}$$

Now solve the equation x=\frac{0.4±\frac{2\sqrt{2}}{5}}{-2} when ± is plus. Add 0.4 to \frac{2\sqrt{2}}{5}.

$$x=\frac{2\sqrt{2}+2}{-2\times 5}$$

Divide \frac{2+2\sqrt{2}}{5} by -2.

$$x=\frac{-\sqrt{2}-1}{5}$$

Now solve the equation x=\frac{0.4±\frac{2\sqrt{2}}{5}}{-2} when ± is minus. Subtract \frac{2\sqrt{2}}{5} from 0.4.

$$x=\frac{2-2\sqrt{2}}{-2\times 5}$$

Divide \frac{2-2\sqrt{2}}{5} by -2.

$$x=\frac{\sqrt{2}-1}{5}$$

The equation is now solved.

$$x=\frac{-\sqrt{2}-1}{5} x=\frac{\sqrt{2}-1}{5}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(0.2-x\right)^{2}.

$$0.04-0.4x+x^{2}=2x^{2}$$

Subtract 2x^{2} from both sides.

$$0.04-0.4x+x^{2}-2x^{2}=0$$

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

$$0.04-0.4x-x^{2}=0$$

Subtract 0.04 from both sides. Anything subtracted from zero gives its negation.

$$-0.4x-x^{2}=-0.04$$

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.

$$-x^{2}-0.4x=-0.04$$

Divide both sides by -1.

$$\frac{-x^{2}-0.4x}{-1}=-\frac{0.04}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{0.4}{-1}\right)x=-\frac{0.04}{-1}$$

Divide -0.4 by -1.

$$x^{2}+0.4x=-\frac{0.04}{-1}$$

Divide -0.04 by -1.

$$x^{2}+0.4x=0.04$$

Divide 0.4, the coefficient of the x term, by 2 to get 0.2. Then add the square of 0.2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+0.4x+0.2^{2}=0.04+0.2^{2}$$

Square 0.2 by squaring both the numerator and the denominator of the fraction.

$$x^{2}+0.4x+0.04=\frac{1+1}{25}$$

Add 0.04 to 0.04 by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+0.4x+0.04=0.08$$

Factor x^{2}+0.4x+0.04. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+0.2\right)^{2}=0.08$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+0.2\right)^{2}}=\sqrt{0.08}$$

Simplify.

$$x+0.2=\frac{\sqrt{2}}{5} x+0.2=-\frac{\sqrt{2}}{5}$$

Subtract 0.2 from both sides of the equation.

$$x=\frac{\sqrt{2}-1}{5} x=\frac{-\sqrt{2}-1}{5}$$