Solve 2-2x=5x^2

$$2 - 2 x = 5 x ^ { 2 }$$

Solve for x

$$x=\frac{\sqrt{11}-1}{5}\approx 0.463324958$$

$$x=\frac{-\sqrt{11}-1}{5}\approx -0.863324958$$

Steps Using the Quadratic Formula

Subtract 5x^{2} from both sides.

$$2-2x-5x^{2}=0$$

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.

$$-5x^{2}-2x+2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, -2 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\left(-5\right)\times 2}}{2\left(-5\right)}$$

Square -2.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\left(-5\right)\times 2}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$x=\frac{-\left(-2\right)±\sqrt{4+20\times 2}}{2\left(-5\right)}$$

Multiply 20 times 2.

$$x=\frac{-\left(-2\right)±\sqrt{4+40}}{2\left(-5\right)}$$

Add 4 to 40.

$$x=\frac{-\left(-2\right)±\sqrt{44}}{2\left(-5\right)}$$

Take the square root of 44.

$$x=\frac{-\left(-2\right)±2\sqrt{11}}{2\left(-5\right)}$$

The opposite of -2 is 2.

$$x=\frac{2±2\sqrt{11}}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{2±2\sqrt{11}}{-10}$$

Now solve the equation x=\frac{2±2\sqrt{11}}{-10} when ± is plus. Add 2 to 2\sqrt{11}.

$$x=\frac{2\sqrt{11}+2}{-10}$$

Divide 2+2\sqrt{11} by -10.

$$x=\frac{-\sqrt{11}-1}{5}$$

Now solve the equation x=\frac{2±2\sqrt{11}}{-10} when ± is minus. Subtract 2\sqrt{11} from 2.

$$x=\frac{2-2\sqrt{11}}{-10}$$

Divide 2-2\sqrt{11} by -10.

$$x=\frac{\sqrt{11}-1}{5}$$

The equation is now solved.

$$x=\frac{-\sqrt{11}-1}{5} x=\frac{\sqrt{11}-1}{5}$$

Steps for Completing the Square

Subtract 5x^{2} from both sides.

$$2-2x-5x^{2}=0$$

Subtract 2 from both sides. Anything subtracted from zero gives its negation.

$$-2x-5x^{2}=-2$$

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.

$$-5x^{2}-2x=-2$$

Divide both sides by -5.

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

Dividing by -5 undoes the multiplication by -5.

$$x^{2}+\left(-\frac{2}{-5}\right)x=-\frac{2}{-5}$$

Divide -2 by -5.

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

Divide -2 by -5.

$$x^{2}+\frac{2}{5}x=\frac{2}{5}$$

Divide \frac{2}{5}, the coefficient of the x term, by 2 to get \frac{1}{5}. Then add the square of \frac{1}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{2}{5}x+\left(\frac{1}{5}\right)^{2}=\frac{2}{5}+\left(\frac{1}{5}\right)^{2}$$

Square \frac{1}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{2}{5}x+\frac{1}{25}=\frac{2}{5}+\frac{1}{25}$$

Add \frac{2}{5} to \frac{1}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{2}{5}x+\frac{1}{25}=\frac{11}{25}$$

Factor x^{2}+\frac{2}{5}x+\frac{1}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{5}\right)^{2}=\frac{11}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{5}\right)^{2}}=\sqrt{\frac{11}{25}}$$

Simplify.

$$x+\frac{1}{5}=\frac{\sqrt{11}}{5} x+\frac{1}{5}=-\frac{\sqrt{11}}{5}$$

Subtract \frac{1}{5} from both sides of the equation.

$$x=\frac{\sqrt{11}-1}{5} x=\frac{-\sqrt{11}-1}{5}$$