Solve -0.4=2x-5x^2

$$- 0.4 = 2 x - 5 x ^ { 2 }$$

Solve for x

$$x=\frac{\sqrt{3}+1}{5}\approx 0.546410162$$

$$x=\frac{1-\sqrt{3}}{5}\approx -0.146410162$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$2x-5x^{2}=-0.4$$

Add 0.4 to both sides.

$$2x-5x^{2}+0.4=0$$

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.

$$-5x^{2}+2x+0.4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, 2 for b, and 0.4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2±\sqrt{2^{2}-4\left(-5\right)\times 0.4}}{2\left(-5\right)}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\left(-5\right)\times 0.4}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$x=\frac{-2±\sqrt{4+20\times 0.4}}{2\left(-5\right)}$$

Multiply 20 times 0.4.

$$x=\frac{-2±\sqrt{4+8}}{2\left(-5\right)}$$

Add 4 to 8.

$$x=\frac{-2±\sqrt{12}}{2\left(-5\right)}$$

Take the square root of 12.

$$x=\frac{-2±2\sqrt{3}}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{-2±2\sqrt{3}}{-10}$$

Now solve the equation x=\frac{-2±2\sqrt{3}}{-10} when ± is plus. Add -2 to 2\sqrt{3}.

$$x=\frac{2\sqrt{3}-2}{-10}$$

Divide -2+2\sqrt{3} by -10.

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

Now solve the equation x=\frac{-2±2\sqrt{3}}{-10} when ± is minus. Subtract 2\sqrt{3} from -2.

$$x=\frac{-2\sqrt{3}-2}{-10}$$

Divide -2-2\sqrt{3} by -10.

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

The equation is now solved.

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

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

$$2x-5x^{2}=-0.4$$

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.

$$-5x^{2}+2x=-0.4$$

Divide both sides by -5.

$$\frac{-5x^{2}+2x}{-5}=-\frac{0.4}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$x^{2}+\frac{2}{-5}x=-\frac{0.4}{-5}$$

Divide 2 by -5.

$$x^{2}-\frac{2}{5}x=-\frac{0.4}{-5}$$

Divide -0.4 by -5.

$$x^{2}-\frac{2}{5}x=0.08$$

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.

$$x^{2}-\frac{2}{5}x+\left(-\frac{1}{5}\right)^{2}=0.08+\left(-\frac{1}{5}\right)^{2}$$

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

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

Add 0.08 to \frac{1}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{2}{5}x+\frac{1}{25}=\frac{3}{25}$$

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}.

$$\left(x-\frac{1}{5}\right)^{2}=\frac{3}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{5}\right)^{2}}=\sqrt{\frac{3}{25}}$$

Simplify.

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

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

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