Solve x/5=2+5x^2

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

Solve for x (complex solution)

$$x=\frac{1+3\sqrt{111}i}{50}\approx 0.02+0.632139225i$$

$$x=\frac{-3\sqrt{111}i+1}{50}\approx 0.02-0.632139225i$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 5.

$$x=10+25x^{2}$$

Subtract 10 from both sides.

$$x-10=25x^{2}$$

Subtract 25x^{2} from both sides.

$$x-10-25x^{2}=0$$

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.

$$-25x^{2}+x-10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -25 for a, 1 for b, and -10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}-4\left(-25\right)\left(-10\right)}}{2\left(-25\right)}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\left(-25\right)\left(-10\right)}}{2\left(-25\right)}$$

Multiply -4 times -25.

$$x=\frac{-1±\sqrt{1+100\left(-10\right)}}{2\left(-25\right)}$$

Multiply 100 times -10.

$$x=\frac{-1±\sqrt{1-1000}}{2\left(-25\right)}$$

Add 1 to -1000.

$$x=\frac{-1±\sqrt{-999}}{2\left(-25\right)}$$

Take the square root of -999.

$$x=\frac{-1±3\sqrt{111}i}{2\left(-25\right)}$$

Multiply 2 times -25.

$$x=\frac{-1±3\sqrt{111}i}{-50}$$

Now solve the equation x=\frac{-1±3\sqrt{111}i}{-50} when ± is plus. Add -1 to 3i\sqrt{111}.

$$x=\frac{-1+3\sqrt{111}i}{-50}$$

Divide -1+3i\sqrt{111} by -50.

$$x=\frac{-3\sqrt{111}i+1}{50}$$

Now solve the equation x=\frac{-1±3\sqrt{111}i}{-50} when ± is minus. Subtract 3i\sqrt{111} from -1.

$$x=\frac{-3\sqrt{111}i-1}{-50}$$

Divide -1-3i\sqrt{111} by -50.

$$x=\frac{1+3\sqrt{111}i}{50}$$

The equation is now solved.

$$x=\frac{-3\sqrt{111}i+1}{50} x=\frac{1+3\sqrt{111}i}{50}$$

Steps for Completing the Square

Multiply both sides of the equation by 5.

$$x=10+25x^{2}$$

Subtract 25x^{2} from both sides.

$$x-25x^{2}=10$$

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.

$$-25x^{2}+x=10$$

Divide both sides by -25.

$$\frac{-25x^{2}+x}{-25}=\frac{10}{-25}$$

Dividing by -25 undoes the multiplication by -25.

$$x^{2}+\frac{1}{-25}x=\frac{10}{-25}$$

Divide 1 by -25.

$$x^{2}-\frac{1}{25}x=\frac{10}{-25}$$

Reduce the fraction \frac{10}{-25} to lowest terms by extracting and canceling out 5.

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

Divide -\frac{1}{25}, the coefficient of the x term, by 2 to get -\frac{1}{50}. Then add the square of -\frac{1}{50} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{25}x+\left(-\frac{1}{50}\right)^{2}=-\frac{2}{5}+\left(-\frac{1}{50}\right)^{2}$$

Square -\frac{1}{50} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{25}x+\frac{1}{2500}=-\frac{2}{5}+\frac{1}{2500}$$

Add -\frac{2}{5} to \frac{1}{2500} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{25}x+\frac{1}{2500}=-\frac{999}{2500}$$

Factor x^{2}-\frac{1}{25}x+\frac{1}{2500}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{50}\right)^{2}=-\frac{999}{2500}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{50}\right)^{2}}=\sqrt{-\frac{999}{2500}}$$

Simplify.

$$x-\frac{1}{50}=\frac{3\sqrt{111}i}{50} x-\frac{1}{50}=-\frac{3\sqrt{111}i}{50}$$

Add \frac{1}{50} to both sides of the equation.

$$x=\frac{1+3\sqrt{111}i}{50} x=\frac{-3\sqrt{111}i+1}{50}$$