Solve 4-5x^2=8x

$$4 - 5 x ^ { 2 } = 8 x$$

Solve for x

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

$$x=-2$$

Steps Using Factoring By Grouping

Subtract 8x from both sides.

$$4-5x^{2}-8x=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-5x^{2}-8x+4=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -5x^{2}+ax+bx+4. To find a and b, set up a system to be solved.

$$a+b=-8 ab=-5\times 4=-20$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -20.

$$1,-20 2,-10 4,-5$$

Calculate the sum for each pair.

$$1-20=-19 2-10=-8 4-5=-1$$

The solution is the pair that gives sum -8.

$$a=2 b=-10$$

Rewrite -5x^{2}-8x+4 as \left(-5x^{2}+2x\right)+\left(-10x+4\right).

$$\left(-5x^{2}+2x\right)+\left(-10x+4\right)$$

Factor out -x in the first and -2 in the second group.

$$-x\left(5x-2\right)-2\left(5x-2\right)$$

Factor out common term 5x-2 by using distributive property.

$$\left(5x-2\right)\left(-x-2\right)$$

To find equation solutions, solve 5x-2=0 and -x-2=0.

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

Steps Using the Quadratic Formula

Subtract 8x from both sides.

$$4-5x^{2}-8x=0$$

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.

$$-5x^{2}-8x+4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, -8 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\left(-5\right)\times 4}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$x=\frac{-\left(-8\right)±\sqrt{64+20\times 4}}{2\left(-5\right)}$$

Multiply 20 times 4.

$$x=\frac{-\left(-8\right)±\sqrt{64+80}}{2\left(-5\right)}$$

Add 64 to 80.

$$x=\frac{-\left(-8\right)±\sqrt{144}}{2\left(-5\right)}$$

Take the square root of 144.

$$x=\frac{-\left(-8\right)±12}{2\left(-5\right)}$$

The opposite of -8 is 8.

$$x=\frac{8±12}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{8±12}{-10}$$

Now solve the equation x=\frac{8±12}{-10} when ± is plus. Add 8 to 12.

$$x=\frac{20}{-10}$$

Divide 20 by -10.

$$x=-2$$

Now solve the equation x=\frac{8±12}{-10} when ± is minus. Subtract 12 from 8.

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

Reduce the fraction \frac{-4}{-10} to lowest terms by extracting and canceling out 2.

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

The equation is now solved.

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

Steps for Completing the Square

Subtract 8x from both sides.

$$4-5x^{2}-8x=0$$

Subtract 4 from both sides. Anything subtracted from zero gives its negation.

$$-5x^{2}-8x=-4$$

Divide both sides by -5.

$$\frac{-5x^{2}-8x}{-5}=-\frac{4}{-5}$$

Dividing by -5 undoes the multiplication by -5.

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

Divide -8 by -5.

$$x^{2}+\frac{8}{5}x=-\frac{4}{-5}$$

Divide -4 by -5.

$$x^{2}+\frac{8}{5}x=\frac{4}{5}$$

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.

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

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

$$x^{2}+\frac{8}{5}x+\frac{16}{25}=\frac{4}{5}+\frac{16}{25}$$

Add \frac{4}{5} to \frac{16}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{8}{5}x+\frac{16}{25}=\frac{36}{25}$$

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}.

$$\left(x+\frac{4}{5}\right)^{2}=\frac{36}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{4}{5}\right)^{2}}=\sqrt{\frac{36}{25}}$$

Simplify.

$$x+\frac{4}{5}=\frac{6}{5} x+\frac{4}{5}=-\frac{6}{5}$$

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

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