Solve x^2-0.24=x

$$x ^ { 2 } - 0.24 = x$$

Solve for x

$$x=1.2$$

$$x=-0.2$$

Steps Using the Quadratic Formula

Subtract x from both sides.

$$x^{2}-0.24-x=0$$

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.

$$x^{2}-x-0.24=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -1 for b, and -0.24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\left(-0.24\right)}}{2}$$

Multiply -4 times -0.24.

$$x=\frac{-\left(-1\right)±\sqrt{1+0.96}}{2}$$

Add 1 to 0.96.

$$x=\frac{-\left(-1\right)±\sqrt{1.96}}{2}$$

Take the square root of 1.96.

$$x=\frac{-\left(-1\right)±\frac{7}{5}}{2}$$

The opposite of -1 is 1.

$$x=\frac{1±\frac{7}{5}}{2}$$

Now solve the equation x=\frac{1±\frac{7}{5}}{2} when ± is plus. Add 1 to \frac{7}{5}.

$$x=\frac{\frac{12}{5}}{2}$$

Divide \frac{12}{5} by 2.

$$x=\frac{6}{5}$$

Now solve the equation x=\frac{1±\frac{7}{5}}{2} when ± is minus. Subtract \frac{7}{5} from 1.

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

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

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

The equation is now solved.

$$x=\frac{6}{5} x=-\frac{1}{5}$$

Steps for Completing the Square

Subtract x from both sides.

$$x^{2}-0.24-x=0$$

Add 0.24 to both sides. Anything plus zero gives itself.

$$x^{2}-x=0.24$$

Divide -1, the coefficient of the x term, by 2 to get -\frac{1}{2}. Then add the square of -\frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-x+\left(-\frac{1}{2}\right)^{2}=0.24+\left(-\frac{1}{2}\right)^{2}$$

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-x+\frac{1}{4}=0.24+\frac{1}{4}$$

Add 0.24 to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-x+\frac{1}{4}=\frac{49}{100}$$

Factor x^{2}-x+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{49}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{49}{100}}$$

Simplify.

$$x-\frac{1}{2}=\frac{7}{10} x-\frac{1}{2}=-\frac{7}{10}$$

Add \frac{1}{2} to both sides of the equation.

$$x=\frac{6}{5} x=-\frac{1}{5}$$