Solve x^2=12-0.2x

$${ x }^{ 2 } =12-0.2x$$

Solve for x

$$x = \frac{\sqrt{1201} - 1}{10} \approx 3.36554469$$

$$x=\frac{-\sqrt{1201}-1}{10}\approx -3.56554469$$

Steps Using the Quadratic Formula

Subtract 12 from both sides.

$$x^{2}-12=-0.2x$$

Add 0.2x to both sides.

$$x^{2}-12+0.2x=0$$

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.

$$x^{2}+0.2x-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0.2 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-0.2±\sqrt{0.2^{2}-4\left(-12\right)}}{2}$$

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

$$x=\frac{-0.2±\sqrt{0.04-4\left(-12\right)}}{2}$$

Multiply -4 times -12.

$$x=\frac{-0.2±\sqrt{0.04+48}}{2}$$

Add 0.04 to 48.

$$x=\frac{-0.2±\sqrt{48.04}}{2}$$

Take the square root of 48.04.

$$x=\frac{-0.2±\frac{\sqrt{1201}}{5}}{2}$$

Now solve the equation x=\frac{-0.2±\frac{\sqrt{1201}}{5}}{2} when ± is plus. Add -0.2 to \frac{\sqrt{1201}}{5}.

$$x=\frac{\sqrt{1201}-1}{2\times 5}$$

Divide \frac{-1+\sqrt{1201}}{5} by 2.

$$x=\frac{\sqrt{1201}-1}{10}$$

Now solve the equation x=\frac{-0.2±\frac{\sqrt{1201}}{5}}{2} when ± is minus. Subtract \frac{\sqrt{1201}}{5} from -0.2.

$$x=\frac{-\sqrt{1201}-1}{2\times 5}$$

Divide \frac{-1-\sqrt{1201}}{5} by 2.

$$x=\frac{-\sqrt{1201}-1}{10}$$

The equation is now solved.

$$x=\frac{\sqrt{1201}-1}{10} x=\frac{-\sqrt{1201}-1}{10}$$

Steps for Completing the Square

Add 0.2x to both sides.

$$x^{2}+0.2x=12$$

Divide 0.2, the coefficient of the x term, by 2 to get 0.1. Then add the square of 0.1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+0.2x+0.1^{2}=12+0.1^{2}$$

Square 0.1 by squaring both the numerator and the denominator of the fraction.

$$x^{2}+0.2x+0.01=12+0.01$$

Add 12 to 0.01.

$$x^{2}+0.2x+0.01=12.01$$

Factor x^{2}+0.2x+0.01. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+0.1\right)^{2}=12.01$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+0.1\right)^{2}}=\sqrt{12.01}$$

Simplify.

$$x+0.1=\frac{\sqrt{1201}}{10} x+0.1=-\frac{\sqrt{1201}}{10}$$

Subtract 0.1 from both sides of the equation.

$$x=\frac{\sqrt{1201}-1}{10} x=\frac{-\sqrt{1201}-1}{10}$$