Solve 0.2x^2-0.1=0.4x

$$0.2 x ^ { 2 } - 0.1 = 0.4 x$$

Solve for x

$$x=\frac{\sqrt{6}}{2}+1\approx 2.224744871$$

$$x=-\frac{\sqrt{6}}{2}+1\approx -0.224744871$$

Steps Using the Quadratic Formula

Subtract 0.4x from both sides.

$$0.2x^{2}-0.1-0.4x=0$$

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.

$$0.2x^{2}-0.4x-0.1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 0.2 for a, -0.4 for b, and -0.1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-0.4\right)±\sqrt{\left(-0.4\right)^{2}-4\times 0.2\left(-0.1\right)}}{2\times 0.2}$$

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

$$x=\frac{-\left(-0.4\right)±\sqrt{0.16-4\times 0.2\left(-0.1\right)}}{2\times 0.2}$$

Multiply -4 times 0.2.

$$x=\frac{-\left(-0.4\right)±\sqrt{0.16-0.8\left(-0.1\right)}}{2\times 0.2}$$

Multiply -0.8 times -0.1 by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-0.4\right)±\sqrt{\frac{4+2}{25}}}{2\times 0.2}$$

Add 0.16 to 0.08 by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-0.4\right)±\sqrt{0.24}}{2\times 0.2}$$

Take the square root of 0.24.

$$x=\frac{-\left(-0.4\right)±\frac{\sqrt{6}}{5}}{2\times 0.2}$$

The opposite of -0.4 is 0.4.

$$x=\frac{0.4±\frac{\sqrt{6}}{5}}{2\times 0.2}$$

Multiply 2 times 0.2.

$$x=\frac{0.4±\frac{\sqrt{6}}{5}}{0.4}$$

Now solve the equation x=\frac{0.4±\frac{\sqrt{6}}{5}}{0.4} when ± is plus. Add 0.4 to \frac{\sqrt{6}}{5}.

$$x=\frac{\sqrt{6}+2}{0.4\times 5}$$

Divide \frac{2+\sqrt{6}}{5} by 0.4 by multiplying \frac{2+\sqrt{6}}{5} by the reciprocal of 0.4.

$$x=\frac{\sqrt{6}}{2}+1$$

Now solve the equation x=\frac{0.4±\frac{\sqrt{6}}{5}}{0.4} when ± is minus. Subtract \frac{\sqrt{6}}{5} from 0.4.

$$x=\frac{2-\sqrt{6}}{0.4\times 5}$$

Divide \frac{2-\sqrt{6}}{5} by 0.4 by multiplying \frac{2-\sqrt{6}}{5} by the reciprocal of 0.4.

$$x=-\frac{\sqrt{6}}{2}+1$$

The equation is now solved.

$$x=\frac{\sqrt{6}}{2}+1 x=-\frac{\sqrt{6}}{2}+1$$

Steps for Completing the Square

Subtract 0.4x from both sides.

$$0.2x^{2}-0.1-0.4x=0$$

Add 0.1 to both sides. Anything plus zero gives itself.

$$0.2x^{2}-0.4x=0.1$$

Multiply both sides by 5.

$$\frac{0.2x^{2}-0.4x}{0.2}=\frac{0.1}{0.2}$$

Dividing by 0.2 undoes the multiplication by 0.2.

$$x^{2}+\left(-\frac{0.4}{0.2}\right)x=\frac{0.1}{0.2}$$

Divide -0.4 by 0.2 by multiplying -0.4 by the reciprocal of 0.2.

$$x^{2}-2x=\frac{0.1}{0.2}$$

Divide 0.1 by 0.2 by multiplying 0.1 by the reciprocal of 0.2.

$$x^{2}-2x=0.5$$

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.

$$x^{2}-2x+1=0.5+1$$

Add 0.5 to 1.

$$x^{2}-2x+1=1.5$$

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}.

$$\left(x-1\right)^{2}=1.5$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-1\right)^{2}}=\sqrt{1.5}$$

Simplify.

$$x-1=\frac{\sqrt{6}}{2} x-1=-\frac{\sqrt{6}}{2}$$

Add 1 to both sides of the equation.

$$x=\frac{\sqrt{6}}{2}+1 x=-\frac{\sqrt{6}}{2}+1$$