Solve x^2+x-2x=8

$$x ^ { 2 } + x - 2 x = 8$$

Solve for x

$$x = \frac{\sqrt{33} + 1}{2} \approx 3.372281323$$

$$x=\frac{1-\sqrt{33}}{2}\approx -2.372281323$$

Steps Using the Quadratic Formula

Combine x and -2x to get -x.

$$x^{2}-x=8$$

Subtract 8 from both sides.

$$x^{2}-x-8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -1 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\left(-8\right)}}{2}$$

Multiply -4 times -8.

$$x=\frac{-\left(-1\right)±\sqrt{1+32}}{2}$$

Add 1 to 32.

$$x=\frac{-\left(-1\right)±\sqrt{33}}{2}$$

The opposite of -1 is 1.

$$x=\frac{1±\sqrt{33}}{2}$$

Now solve the equation x=\frac{1±\sqrt{33}}{2} when ± is plus. Add 1 to \sqrt{33}.

$$x=\frac{\sqrt{33}+1}{2}$$

Now solve the equation x=\frac{1±\sqrt{33}}{2} when ± is minus. Subtract \sqrt{33} from 1.

$$x=\frac{1-\sqrt{33}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{33}+1}{2} x=\frac{1-\sqrt{33}}{2}$$

Steps for Completing the Square

Combine x and -2x to get -x.

$$x^{2}-x=8$$

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.

$$x^{2}-x+\left(-\frac{1}{2}\right)^{2}=8+\left(-\frac{1}{2}\right)^{2}$$

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

$$x^{2}-x+\frac{1}{4}=8+\frac{1}{4}$$

Add 8 to \frac{1}{4}.

$$x^{2}-x+\frac{1}{4}=\frac{33}{4}$$

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}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{33}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{33}{4}}$$

Simplify.

$$x-\frac{1}{2}=\frac{\sqrt{33}}{2} x-\frac{1}{2}=-\frac{\sqrt{33}}{2}$$

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

$$x=\frac{\sqrt{33}+1}{2} x=\frac{1-\sqrt{33}}{2}$$