Solve (20-3x)(8-2x)=56

$$(20-3x)(8-2x)=56$$

Solve for x

$$x=2$$

$$x = \frac{26}{3} = 8\frac{2}{3} \approx 8.666666667$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 20-3x by 8-2x and combine like terms.

$$160-64x+6x^{2}=56$$

Subtract 56 from both sides.

$$160-64x+6x^{2}-56=0$$

Subtract 56 from 160 to get 104.

$$104-64x+6x^{2}=0$$

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.

$$6x^{2}-64x+104=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, -64 for b, and 104 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-64\right)±\sqrt{\left(-64\right)^{2}-4\times 6\times 104}}{2\times 6}$$

Square -64.

$$x=\frac{-\left(-64\right)±\sqrt{4096-4\times 6\times 104}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-\left(-64\right)±\sqrt{4096-24\times 104}}{2\times 6}$$

Multiply -24 times 104.

$$x=\frac{-\left(-64\right)±\sqrt{4096-2496}}{2\times 6}$$

Add 4096 to -2496.

$$x=\frac{-\left(-64\right)±\sqrt{1600}}{2\times 6}$$

Take the square root of 1600.

$$x=\frac{-\left(-64\right)±40}{2\times 6}$$

The opposite of -64 is 64.

$$x=\frac{64±40}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{64±40}{12}$$

Now solve the equation x=\frac{64±40}{12} when ± is plus. Add 64 to 40.

$$x=\frac{104}{12}$$

Reduce the fraction \frac{104}{12} to lowest terms by extracting and canceling out 4.

$$x=\frac{26}{3}$$

Now solve the equation x=\frac{64±40}{12} when ± is minus. Subtract 40 from 64.

$$x=\frac{24}{12}$$

Divide 24 by 12.

$$x=2$$

The equation is now solved.

$$x=\frac{26}{3} x=2$$

Steps for Completing the Square

Use the distributive property to multiply 20-3x by 8-2x and combine like terms.

$$160-64x+6x^{2}=56$$

Subtract 160 from both sides.

$$-64x+6x^{2}=56-160$$

Subtract 160 from 56 to get -104.

$$-64x+6x^{2}=-104$$

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.

$$6x^{2}-64x=-104$$

Divide both sides by 6.

$$\frac{6x^{2}-64x}{6}=-\frac{104}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$x^{2}+\left(-\frac{64}{6}\right)x=-\frac{104}{6}$$

Reduce the fraction \frac{-64}{6} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{32}{3}x=-\frac{104}{6}$$

Reduce the fraction \frac{-104}{6} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{32}{3}x=-\frac{52}{3}$$

Divide -\frac{32}{3}, the coefficient of the x term, by 2 to get -\frac{16}{3}. Then add the square of -\frac{16}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{32}{3}x+\left(-\frac{16}{3}\right)^{2}=-\frac{52}{3}+\left(-\frac{16}{3}\right)^{2}$$

Square -\frac{16}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{32}{3}x+\frac{256}{9}=-\frac{52}{3}+\frac{256}{9}$$

Add -\frac{52}{3} to \frac{256}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{32}{3}x+\frac{256}{9}=\frac{100}{9}$$

Factor x^{2}-\frac{32}{3}x+\frac{256}{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}.

$$\left(x-\frac{16}{3}\right)^{2}=\frac{100}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{16}{3}\right)^{2}}=\sqrt{\frac{100}{9}}$$

Simplify.

$$x-\frac{16}{3}=\frac{10}{3} x-\frac{16}{3}=-\frac{10}{3}$$

Add \frac{16}{3} to both sides of the equation.

$$x=\frac{26}{3} x=2$$