Solve (20-3x)(4-x)=28

$$(20-3x)(4-x)=28$$

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 4-x and combine like terms.

$$80-32x+3x^{2}=28$$

Subtract 28 from both sides.

$$80-32x+3x^{2}-28=0$$

Subtract 28 from 80 to get 52.

$$52-32x+3x^{2}=0$$

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.

$$3x^{2}-32x+52=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -32 for b, and 52 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-32\right)±\sqrt{\left(-32\right)^{2}-4\times 3\times 52}}{2\times 3}$$

Square -32.

$$x=\frac{-\left(-32\right)±\sqrt{1024-4\times 3\times 52}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-32\right)±\sqrt{1024-12\times 52}}{2\times 3}$$

Multiply -12 times 52.

$$x=\frac{-\left(-32\right)±\sqrt{1024-624}}{2\times 3}$$

Add 1024 to -624.

$$x=\frac{-\left(-32\right)±\sqrt{400}}{2\times 3}$$

Take the square root of 400.

$$x=\frac{-\left(-32\right)±20}{2\times 3}$$

The opposite of -32 is 32.

$$x=\frac{32±20}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{32±20}{6}$$

Now solve the equation x=\frac{32±20}{6} when ± is plus. Add 32 to 20.

$$x=\frac{52}{6}$$

Reduce the fraction \frac{52}{6} to lowest terms by extracting and canceling out 2.

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

Now solve the equation x=\frac{32±20}{6} when ± is minus. Subtract 20 from 32.

$$x=\frac{12}{6}$$

Divide 12 by 6.

$$x=2$$

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 4-x and combine like terms.

$$80-32x+3x^{2}=28$$

Subtract 80 from both sides.

$$-32x+3x^{2}=28-80$$

Subtract 80 from 28 to get -52.

$$-32x+3x^{2}=-52$$

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.

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

Divide both sides by 3.

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

Dividing by 3 undoes the multiplication by 3.

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

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.

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