Solve 16*20=4/10(16-x)(20-x)

$$16 \times 20 = \frac { 4 } { 10 } ( 16 - x ) ( 20 - x )$$

Solve for x

$$x=18-2\sqrt{201}\approx -10.354893758$$

$$x=2\sqrt{201}+18\approx 46.354893758$$

Steps Using the Quadratic Formula

Multiply 16 and 20 to get 320.

$$320=\frac{4}{10}\left(16-x\right)\left(20-x\right)$$

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

$$320=\frac{2}{5}\left(16-x\right)\left(20-x\right)$$

Use the distributive property to multiply \frac{2}{5} by 16-x.

$$320=\left(\frac{32}{5}-\frac{2}{5}x\right)\left(20-x\right)$$

Use the distributive property to multiply \frac{32}{5}-\frac{2}{5}x by 20-x and combine like terms.

$$320=128-\frac{72}{5}x+\frac{2}{5}x^{2}$$

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

$$128-\frac{72}{5}x+\frac{2}{5}x^{2}=320$$

Subtract 320 from both sides.

$$128-\frac{72}{5}x+\frac{2}{5}x^{2}-320=0$$

Subtract 320 from 128 to get -192.

$$-192-\frac{72}{5}x+\frac{2}{5}x^{2}=0$$

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.

$$\frac{2}{5}x^{2}-\frac{72}{5}x-192=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{2}{5} for a, -\frac{72}{5} for b, and -192 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-\frac{72}{5}\right)±\sqrt{\left(-\frac{72}{5}\right)^{2}-4\times \frac{2}{5}\left(-192\right)}}{2\times \frac{2}{5}}$$

Square -\frac{72}{5} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\left(-\frac{72}{5}\right)±\sqrt{\frac{5184}{25}-4\times \frac{2}{5}\left(-192\right)}}{2\times \frac{2}{5}}$$

Multiply -4 times \frac{2}{5}.

$$x=\frac{-\left(-\frac{72}{5}\right)±\sqrt{\frac{5184}{25}-\frac{8}{5}\left(-192\right)}}{2\times \frac{2}{5}}$$

Multiply -\frac{8}{5} times -192.

$$x=\frac{-\left(-\frac{72}{5}\right)±\sqrt{\frac{5184}{25}+\frac{1536}{5}}}{2\times \frac{2}{5}}$$

Add \frac{5184}{25} to \frac{1536}{5} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-\frac{72}{5}\right)±\sqrt{\frac{12864}{25}}}{2\times \frac{2}{5}}$$

Take the square root of \frac{12864}{25}.

$$x=\frac{-\left(-\frac{72}{5}\right)±\frac{8\sqrt{201}}{5}}{2\times \frac{2}{5}}$$

The opposite of -\frac{72}{5} is \frac{72}{5}.

$$x=\frac{\frac{72}{5}±\frac{8\sqrt{201}}{5}}{2\times \frac{2}{5}}$$

Multiply 2 times \frac{2}{5}.

$$x=\frac{\frac{72}{5}±\frac{8\sqrt{201}}{5}}{\frac{4}{5}}$$

Now solve the equation x=\frac{\frac{72}{5}±\frac{8\sqrt{201}}{5}}{\frac{4}{5}} when ± is plus. Add \frac{72}{5} to \frac{8\sqrt{201}}{5}.

$$x=\frac{8\sqrt{201}+72}{\frac{4}{5}\times 5}$$

Divide \frac{72+8\sqrt{201}}{5} by \frac{4}{5} by multiplying \frac{72+8\sqrt{201}}{5} by the reciprocal of \frac{4}{5}.

$$x=2\sqrt{201}+18$$

Now solve the equation x=\frac{\frac{72}{5}±\frac{8\sqrt{201}}{5}}{\frac{4}{5}} when ± is minus. Subtract \frac{8\sqrt{201}}{5} from \frac{72}{5}.

$$x=\frac{72-8\sqrt{201}}{\frac{4}{5}\times 5}$$

Divide \frac{72-8\sqrt{201}}{5} by \frac{4}{5} by multiplying \frac{72-8\sqrt{201}}{5} by the reciprocal of \frac{4}{5}.

$$x=18-2\sqrt{201}$$

The equation is now solved.

$$x=2\sqrt{201}+18 x=18-2\sqrt{201}$$

Steps for Completing the Square

Multiply 16 and 20 to get 320.

$$320=\frac{4}{10}\left(16-x\right)\left(20-x\right)$$

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

$$320=\frac{2}{5}\left(16-x\right)\left(20-x\right)$$

Use the distributive property to multiply \frac{2}{5} by 16-x.

$$320=\left(\frac{32}{5}-\frac{2}{5}x\right)\left(20-x\right)$$

Use the distributive property to multiply \frac{32}{5}-\frac{2}{5}x by 20-x and combine like terms.

$$320=128-\frac{72}{5}x+\frac{2}{5}x^{2}$$

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

$$128-\frac{72}{5}x+\frac{2}{5}x^{2}=320$$

Subtract 128 from both sides.

$$-\frac{72}{5}x+\frac{2}{5}x^{2}=320-128$$

Subtract 128 from 320 to get 192.

$$-\frac{72}{5}x+\frac{2}{5}x^{2}=192$$

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.

$$\frac{2}{5}x^{2}-\frac{72}{5}x=192$$

Divide both sides of the equation by \frac{2}{5}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{2}{5}x^{2}-\frac{72}{5}x}{\frac{2}{5}}=\frac{192}{\frac{2}{5}}$$

Dividing by \frac{2}{5} undoes the multiplication by \frac{2}{5}.

$$x^{2}+\left(-\frac{\frac{72}{5}}{\frac{2}{5}}\right)x=\frac{192}{\frac{2}{5}}$$

Divide -\frac{72}{5} by \frac{2}{5} by multiplying -\frac{72}{5} by the reciprocal of \frac{2}{5}.

$$x^{2}-36x=\frac{192}{\frac{2}{5}}$$

Divide 192 by \frac{2}{5} by multiplying 192 by the reciprocal of \frac{2}{5}.

$$x^{2}-36x=480$$

Divide -36, the coefficient of the x term, by 2 to get -18. Then add the square of -18 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-36x+\left(-18\right)^{2}=480+\left(-18\right)^{2}$$

Square -18.

$$x^{2}-36x+324=480+324$$

Add 480 to 324.

$$x^{2}-36x+324=804$$

Factor x^{2}-36x+324. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-18\right)^{2}=804$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-18\right)^{2}}=\sqrt{804}$$

Simplify.

$$x-18=2\sqrt{201} x-18=-2\sqrt{201}$$

Add 18 to both sides of the equation.

$$x=2\sqrt{201}+18 x=18-2\sqrt{201}$$