Solve 8(2-5x)=25x^2

$$8 ( 2 - 5 x ) = 25 x ^ { 2 }$$

Solve for x

$$x=\frac{4\sqrt{2}-4}{5}\approx 0.33137085$$

$$x=\frac{-4\sqrt{2}-4}{5}\approx -1.93137085$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 8 by 2-5x.

$$16-40x=25x^{2}$$

Subtract 25x^{2} from both sides.

$$16-40x-25x^{2}=0$$

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.

$$-25x^{2}-40x+16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -25 for a, -40 for b, and 16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-40\right)±\sqrt{\left(-40\right)^{2}-4\left(-25\right)\times 16}}{2\left(-25\right)}$$

Square -40.

$$x=\frac{-\left(-40\right)±\sqrt{1600-4\left(-25\right)\times 16}}{2\left(-25\right)}$$

Multiply -4 times -25.

$$x=\frac{-\left(-40\right)±\sqrt{1600+100\times 16}}{2\left(-25\right)}$$

Multiply 100 times 16.

$$x=\frac{-\left(-40\right)±\sqrt{1600+1600}}{2\left(-25\right)}$$

Add 1600 to 1600.

$$x=\frac{-\left(-40\right)±\sqrt{3200}}{2\left(-25\right)}$$

Take the square root of 3200.

$$x=\frac{-\left(-40\right)±40\sqrt{2}}{2\left(-25\right)}$$

The opposite of -40 is 40.

$$x=\frac{40±40\sqrt{2}}{2\left(-25\right)}$$

Multiply 2 times -25.

$$x=\frac{40±40\sqrt{2}}{-50}$$

Now solve the equation x=\frac{40±40\sqrt{2}}{-50} when ± is plus. Add 40 to 40\sqrt{2}.

$$x=\frac{40\sqrt{2}+40}{-50}$$

Divide 40+40\sqrt{2} by -50.

$$x=\frac{-4\sqrt{2}-4}{5}$$

Now solve the equation x=\frac{40±40\sqrt{2}}{-50} when ± is minus. Subtract 40\sqrt{2} from 40.

$$x=\frac{40-40\sqrt{2}}{-50}$$

Divide 40-40\sqrt{2} by -50.

$$x=\frac{4\sqrt{2}-4}{5}$$

The equation is now solved.

$$x=\frac{-4\sqrt{2}-4}{5} x=\frac{4\sqrt{2}-4}{5}$$

Steps for Completing the Square

Use the distributive property to multiply 8 by 2-5x.

$$16-40x=25x^{2}$$

Subtract 25x^{2} from both sides.

$$16-40x-25x^{2}=0$$

Subtract 16 from both sides. Anything subtracted from zero gives its negation.

$$-40x-25x^{2}=-16$$

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.

$$-25x^{2}-40x=-16$$

Divide both sides by -25.

$$\frac{-25x^{2}-40x}{-25}=-\frac{16}{-25}$$

Dividing by -25 undoes the multiplication by -25.

$$x^{2}+\left(-\frac{40}{-25}\right)x=-\frac{16}{-25}$$

Reduce the fraction \frac{-40}{-25} to lowest terms by extracting and canceling out 5.

$$x^{2}+\frac{8}{5}x=-\frac{16}{-25}$$

Divide -16 by -25.

$$x^{2}+\frac{8}{5}x=\frac{16}{25}$$

Divide \frac{8}{5}, the coefficient of the x term, by 2 to get \frac{4}{5}. Then add the square of \frac{4}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{8}{5}x+\left(\frac{4}{5}\right)^{2}=\frac{16}{25}+\left(\frac{4}{5}\right)^{2}$$

Square \frac{4}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{8}{5}x+\frac{16}{25}=\frac{16+16}{25}$$

Add \frac{16}{25} to \frac{16}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{8}{5}x+\frac{16}{25}=\frac{32}{25}$$

Factor x^{2}+\frac{8}{5}x+\frac{16}{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}.

$$\left(x+\frac{4}{5}\right)^{2}=\frac{32}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{4}{5}\right)^{2}}=\sqrt{\frac{32}{25}}$$

Simplify.

$$x+\frac{4}{5}=\frac{4\sqrt{2}}{5} x+\frac{4}{5}=-\frac{4\sqrt{2}}{5}$$

Subtract \frac{4}{5} from both sides of the equation.

$$x=\frac{4\sqrt{2}-4}{5} x=\frac{-4\sqrt{2}-4}{5}$$