Solve (6-x)(8-2x)=24

$$(6-x)(8-2x)=24$$

Solve for x

$$x=\sqrt{13}+5\approx 8.605551275$$

$$x=5-\sqrt{13}\approx 1.394448725$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 6-x by 8-2x and combine like terms.

$$48-20x+2x^{2}=24$$

Subtract 24 from both sides.

$$48-20x+2x^{2}-24=0$$

Subtract 24 from 48 to get 24.

$$24-20x+2x^{2}=0$$

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.

$$2x^{2}-20x+24=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -20 for b, and 24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 2\times 24}}{2\times 2}$$

Square -20.

$$x=\frac{-\left(-20\right)±\sqrt{400-4\times 2\times 24}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-20\right)±\sqrt{400-8\times 24}}{2\times 2}$$

Multiply -8 times 24.

$$x=\frac{-\left(-20\right)±\sqrt{400-192}}{2\times 2}$$

Add 400 to -192.

$$x=\frac{-\left(-20\right)±\sqrt{208}}{2\times 2}$$

Take the square root of 208.

$$x=\frac{-\left(-20\right)±4\sqrt{13}}{2\times 2}$$

The opposite of -20 is 20.

$$x=\frac{20±4\sqrt{13}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{20±4\sqrt{13}}{4}$$

Now solve the equation x=\frac{20±4\sqrt{13}}{4} when ± is plus. Add 20 to 4\sqrt{13}.

$$x=\frac{4\sqrt{13}+20}{4}$$

Divide 20+4\sqrt{13} by 4.

$$x=\sqrt{13}+5$$

Now solve the equation x=\frac{20±4\sqrt{13}}{4} when ± is minus. Subtract 4\sqrt{13} from 20.

$$x=\frac{20-4\sqrt{13}}{4}$$

Divide 20-4\sqrt{13} by 4.

$$x=5-\sqrt{13}$$

The equation is now solved.

$$x=\sqrt{13}+5 x=5-\sqrt{13}$$

Steps for Completing the Square

Use the distributive property to multiply 6-x by 8-2x and combine like terms.

$$48-20x+2x^{2}=24$$

Subtract 48 from both sides.

$$-20x+2x^{2}=24-48$$

Subtract 48 from 24 to get -24.

$$-20x+2x^{2}=-24$$

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.

$$2x^{2}-20x=-24$$

Divide both sides by 2.

$$\frac{2x^{2}-20x}{2}=-\frac{24}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{20}{2}\right)x=-\frac{24}{2}$$

Divide -20 by 2.

$$x^{2}-10x=-\frac{24}{2}$$

Divide -24 by 2.

$$x^{2}-10x=-12$$

Divide -10, the coefficient of the x term, by 2 to get -5. Then add the square of -5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-10x+\left(-5\right)^{2}=-12+\left(-5\right)^{2}$$

Square -5.

$$x^{2}-10x+25=-12+25$$

Add -12 to 25.

$$x^{2}-10x+25=13$$

Factor x^{2}-10x+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}.

$$\left(x-5\right)^{2}=13$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{13}$$

Simplify.

$$x-5=\sqrt{13} x-5=-\sqrt{13}$$

Add 5 to both sides of the equation.

$$x=\sqrt{13}+5 x=5-\sqrt{13}$$