Solve 2x=5x^2+1

$$2x=5 { x }^{ 2 } +1$$

Solve for x (complex solution)

$$x=\frac{1}{5}-\frac{2}{5}i=0.2-0.4i$$

$$x=\frac{1}{5}+\frac{2}{5}i=0.2+0.4i$$

Steps Using the Quadratic Formula

Subtract 5x^{2} from both sides.

$$2x-5x^{2}=1$$

Subtract 1 from both sides.

$$2x-5x^{2}-1=0$$

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.

$$-5x^{2}+2x-1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, 2 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2±\sqrt{2^{2}-4\left(-5\right)\left(-1\right)}}{2\left(-5\right)}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\left(-5\right)\left(-1\right)}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$x=\frac{-2±\sqrt{4+20\left(-1\right)}}{2\left(-5\right)}$$

Multiply 20 times -1.

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

Add 4 to -20.

$$x=\frac{-2±\sqrt{-16}}{2\left(-5\right)}$$

Take the square root of -16.

$$x=\frac{-2±4i}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{-2±4i}{-10}$$

Now solve the equation x=\frac{-2±4i}{-10} when ± is plus. Add -2 to 4i.

$$x=\frac{-2+4i}{-10}$$

Divide -2+4i by -10.

$$x=\frac{1}{5}-\frac{2}{5}i$$

Now solve the equation x=\frac{-2±4i}{-10} when ± is minus. Subtract 4i from -2.

$$x=\frac{-2-4i}{-10}$$

Divide -2-4i by -10.

$$x=\frac{1}{5}+\frac{2}{5}i$$

The equation is now solved.

$$x=\frac{1}{5}-\frac{2}{5}i x=\frac{1}{5}+\frac{2}{5}i$$

Steps for Completing the Square

Subtract 5x^{2} from both sides.

$$2x-5x^{2}=1$$

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.

$$-5x^{2}+2x=1$$

Divide both sides by -5.

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

Dividing by -5 undoes the multiplication by -5.

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

Divide 2 by -5.

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

Divide 1 by -5.

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

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.

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

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

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

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

$$x^{2}-\frac{2}{5}x+\frac{1}{25}=-\frac{4}{25}$$

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}.

$$\left(x-\frac{1}{5}\right)^{2}=-\frac{4}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{5}\right)^{2}}=\sqrt{-\frac{4}{25}}$$

Simplify.

$$x-\frac{1}{5}=\frac{2}{5}i x-\frac{1}{5}=-\frac{2}{5}i$$

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

$$x=\frac{1}{5}+\frac{2}{5}i x=\frac{1}{5}-\frac{2}{5}i$$