Solve 5x^2=2x-2

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

Solve for x (complex solution)

$$x=\frac{1}{5}+\frac{3}{5}i=0.2+0.6i$$

$$x=\frac{1}{5}-\frac{3}{5}i=0.2-0.6i$$

Steps Using the Quadratic Formula

Subtract 2x from both sides.

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

Add 2 to both sides.

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

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}.

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

Square -2.

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

Multiply -4 times 5.

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

Multiply -20 times 2.

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

Add 4 to -40.

$$x=\frac{-\left(-2\right)±\sqrt{-36}}{2\times 5}$$

Take the square root of -36.

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

The opposite of -2 is 2.

$$x=\frac{2±6i}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{2±6i}{10}$$

Now solve the equation x=\frac{2±6i}{10} when ± is plus. Add 2 to 6i.

$$x=\frac{2+6i}{10}$$

Divide 2+6i by 10.

$$x=\frac{1}{5}+\frac{3}{5}i$$

Now solve the equation x=\frac{2±6i}{10} when ± is minus. Subtract 6i from 2.

$$x=\frac{2-6i}{10}$$

Divide 2-6i by 10.

$$x=\frac{1}{5}-\frac{3}{5}i$$

The equation is now solved.

$$x=\frac{1}{5}+\frac{3}{5}i x=\frac{1}{5}-\frac{3}{5}i$$

Steps for Completing the Square

Subtract 2x from both sides.

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

Divide both sides by 5.

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

Dividing by 5 undoes the multiplication by 5.

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

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.

$$x^{2}-\frac{2}{5}x+\frac{1}{25}=-\frac{9}{25}$$

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}.

$$\left(x-\frac{1}{5}\right)^{2}=-\frac{9}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{5}\right)^{2}}=\sqrt{-\frac{9}{25}}$$

Simplify.

$$x-\frac{1}{5}=\frac{3}{5}i x-\frac{1}{5}=-\frac{3}{5}i$$

Add \frac{1}{5} to both sides of the equation.

$$x=\frac{1}{5}+\frac{3}{5}i x=\frac{1}{5}-\frac{3}{5}i$$