Solve (20-3x)(12-x)=112

$$( 20 - 3 x ) ( 12 - x ) = 112$$

Solve for x

$$x = \frac{8}{3} = 2\frac{2}{3} \approx 2.666666667$$

$$x=16$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 20-3x by 12-x and combine like terms.

$$240-56x+3x^{2}=112$$

Subtract 112 from both sides.

$$240-56x+3x^{2}-112=0$$

Subtract 112 from 240 to get 128.

$$128-56x+3x^{2}=0$$

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.

$$3x^{2}-56x+128=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -56 for b, and 128 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-56\right)±\sqrt{\left(-56\right)^{2}-4\times 3\times 128}}{2\times 3}$$

Square -56.

$$x=\frac{-\left(-56\right)±\sqrt{3136-4\times 3\times 128}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-56\right)±\sqrt{3136-12\times 128}}{2\times 3}$$

Multiply -12 times 128.

$$x=\frac{-\left(-56\right)±\sqrt{3136-1536}}{2\times 3}$$

Add 3136 to -1536.

$$x=\frac{-\left(-56\right)±\sqrt{1600}}{2\times 3}$$

Take the square root of 1600.

$$x=\frac{-\left(-56\right)±40}{2\times 3}$$

The opposite of -56 is 56.

$$x=\frac{56±40}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{56±40}{6}$$

Now solve the equation x=\frac{56±40}{6} when ± is plus. Add 56 to 40.

$$x=\frac{96}{6}$$

Divide 96 by 6.

$$x=16$$

Now solve the equation x=\frac{56±40}{6} when ± is minus. Subtract 40 from 56.

$$x=\frac{16}{6}$$

Reduce the fraction \frac{16}{6} to lowest terms by extracting and canceling out 2.

$$x=\frac{8}{3}$$

The equation is now solved.

$$x=16 x=\frac{8}{3}$$

Steps for Completing the Square

Use the distributive property to multiply 20-3x by 12-x and combine like terms.

$$240-56x+3x^{2}=112$$

Subtract 240 from both sides.

$$-56x+3x^{2}=112-240$$

Subtract 240 from 112 to get -128.

$$-56x+3x^{2}=-128$$

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.

$$3x^{2}-56x=-128$$

Divide both sides by 3.

$$\frac{3x^{2}-56x}{3}=-\frac{128}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{56}{3}x=-\frac{128}{3}$$

Divide -\frac{56}{3}, the coefficient of the x term, by 2 to get -\frac{28}{3}. Then add the square of -\frac{28}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{56}{3}x+\left(-\frac{28}{3}\right)^{2}=-\frac{128}{3}+\left(-\frac{28}{3}\right)^{2}$$

Square -\frac{28}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{56}{3}x+\frac{784}{9}=-\frac{128}{3}+\frac{784}{9}$$

Add -\frac{128}{3} to \frac{784}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{56}{3}x+\frac{784}{9}=\frac{400}{9}$$

Factor x^{2}-\frac{56}{3}x+\frac{784}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{28}{3}\right)^{2}=\frac{400}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{28}{3}\right)^{2}}=\sqrt{\frac{400}{9}}$$

Simplify.

$$x-\frac{28}{3}=\frac{20}{3} x-\frac{28}{3}=-\frac{20}{3}$$

Add \frac{28}{3} to both sides of the equation.

$$x=16 x=\frac{8}{3}$$