Solve (32-2x)x=120

$$(32-2x)x=120$$

Solve for x

$$x=6$$

$$x=10$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 32-2x by x.

$$32x-2x^{2}=120$$

Subtract 120 from both sides.

$$32x-2x^{2}-120=0$$

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.

$$-2x^{2}+32x-120=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 32 for b, and -120 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-32±\sqrt{32^{2}-4\left(-2\right)\left(-120\right)}}{2\left(-2\right)}$$

Square 32.

$$x=\frac{-32±\sqrt{1024-4\left(-2\right)\left(-120\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-32±\sqrt{1024+8\left(-120\right)}}{2\left(-2\right)}$$

Multiply 8 times -120.

$$x=\frac{-32±\sqrt{1024-960}}{2\left(-2\right)}$$

Add 1024 to -960.

$$x=\frac{-32±\sqrt{64}}{2\left(-2\right)}$$

Take the square root of 64.

$$x=\frac{-32±8}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-32±8}{-4}$$

Now solve the equation x=\frac{-32±8}{-4} when ± is plus. Add -32 to 8.

$$x=-\frac{24}{-4}$$

Divide -24 by -4.

$$x=6$$

Now solve the equation x=\frac{-32±8}{-4} when ± is minus. Subtract 8 from -32.

$$x=-\frac{40}{-4}$$

Divide -40 by -4.

$$x=10$$

The equation is now solved.

$$x=6 x=10$$

Steps for Completing the Square

Use the distributive property to multiply 32-2x by x.

$$32x-2x^{2}=120$$

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.

$$-2x^{2}+32x=120$$

Divide both sides by -2.

$$\frac{-2x^{2}+32x}{-2}=\frac{120}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\frac{32}{-2}x=\frac{120}{-2}$$

Divide 32 by -2.

$$x^{2}-16x=\frac{120}{-2}$$

Divide 120 by -2.

$$x^{2}-16x=-60$$

Divide -16, the coefficient of the x term, by 2 to get -8. Then add the square of -8 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-16x+\left(-8\right)^{2}=-60+\left(-8\right)^{2}$$

Square -8.

$$x^{2}-16x+64=-60+64$$

Add -60 to 64.

$$x^{2}-16x+64=4$$

Factor x^{2}-16x+64. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-8\right)^{2}=4$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-8\right)^{2}}=\sqrt{4}$$

Simplify.

$$x-8=2 x-8=-2$$

Add 8 to both sides of the equation.

$$x=10 x=6$$