Solve 20x^2=32+72x

$$20 { x }^{ 2 } =32+72x$$

Solve for x

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

$$x=4$$

Steps Using Factoring By Grouping

Subtract 32 from both sides.

$$20x^{2}-32=72x$$

Subtract 72x from both sides.

$$20x^{2}-32-72x=0$$

Divide both sides by 4.

$$5x^{2}-8-18x=0$$

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

$$5x^{2}-18x-8=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5x^{2}+ax+bx-8. To find a and b, set up a system to be solved.

$$a+b=-18 ab=5\left(-8\right)=-40$$

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 -40.

$$1,-40 2,-20 4,-10 5,-8$$

Calculate the sum for each pair.

$$1-40=-39 2-20=-18 4-10=-6 5-8=-3$$

The solution is the pair that gives sum -18.

$$a=-20 b=2$$

Rewrite 5x^{2}-18x-8 as \left(5x^{2}-20x\right)+\left(2x-8\right).

$$\left(5x^{2}-20x\right)+\left(2x-8\right)$$

Factor out 5x in the first and 2 in the second group.

$$5x\left(x-4\right)+2\left(x-4\right)$$

Factor out common term x-4 by using distributive property.

$$\left(x-4\right)\left(5x+2\right)$$

To find equation solutions, solve x-4=0 and 5x+2=0.

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

Steps Using the Quadratic Formula

Subtract 32 from both sides.

$$20x^{2}-32=72x$$

Subtract 72x from both sides.

$$20x^{2}-32-72x=0$$

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.

$$20x^{2}-72x-32=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 20 for a, -72 for b, and -32 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-72\right)±\sqrt{\left(-72\right)^{2}-4\times 20\left(-32\right)}}{2\times 20}$$

Square -72.

$$x=\frac{-\left(-72\right)±\sqrt{5184-4\times 20\left(-32\right)}}{2\times 20}$$

Multiply -4 times 20.

$$x=\frac{-\left(-72\right)±\sqrt{5184-80\left(-32\right)}}{2\times 20}$$

Multiply -80 times -32.

$$x=\frac{-\left(-72\right)±\sqrt{5184+2560}}{2\times 20}$$

Add 5184 to 2560.

$$x=\frac{-\left(-72\right)±\sqrt{7744}}{2\times 20}$$

Take the square root of 7744.

$$x=\frac{-\left(-72\right)±88}{2\times 20}$$

The opposite of -72 is 72.

$$x=\frac{72±88}{2\times 20}$$

Multiply 2 times 20.

$$x=\frac{72±88}{40}$$

Now solve the equation x=\frac{72±88}{40} when ± is plus. Add 72 to 88.

$$x=\frac{160}{40}$$

Divide 160 by 40.

$$x=4$$

Now solve the equation x=\frac{72±88}{40} when ± is minus. Subtract 88 from 72.

$$x=-\frac{16}{40}$$

Reduce the fraction \frac{-16}{40} to lowest terms by extracting and canceling out 8.

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

The equation is now solved.

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

Steps for Completing the Square

Subtract 72x from both sides.

$$20x^{2}-72x=32$$

Divide both sides by 20.

$$\frac{20x^{2}-72x}{20}=\frac{32}{20}$$

Dividing by 20 undoes the multiplication by 20.

$$x^{2}+\left(-\frac{72}{20}\right)x=\frac{32}{20}$$

Reduce the fraction \frac{-72}{20} to lowest terms by extracting and canceling out 4.

$$x^{2}-\frac{18}{5}x=\frac{32}{20}$$

Reduce the fraction \frac{32}{20} to lowest terms by extracting and canceling out 4.

$$x^{2}-\frac{18}{5}x=\frac{8}{5}$$

Divide -\frac{18}{5}, the coefficient of the x term, by 2 to get -\frac{9}{5}. Then add the square of -\frac{9}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{18}{5}x+\left(-\frac{9}{5}\right)^{2}=\frac{8}{5}+\left(-\frac{9}{5}\right)^{2}$$

Square -\frac{9}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{18}{5}x+\frac{81}{25}=\frac{8}{5}+\frac{81}{25}$$

Add \frac{8}{5} to \frac{81}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{18}{5}x+\frac{81}{25}=\frac{121}{25}$$

Factor x^{2}-\frac{18}{5}x+\frac{81}{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}.

$$\left(x-\frac{9}{5}\right)^{2}=\frac{121}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{9}{5}\right)^{2}}=\sqrt{\frac{121}{25}}$$

Simplify.

$$x-\frac{9}{5}=\frac{11}{5} x-\frac{9}{5}=-\frac{11}{5}$$

Add \frac{9}{5} to both sides of the equation.

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