Solve (32*x)(20-x)=540

$$( 32 \cdot x ) ( 20 - x ) = 540$$

Solve for x

$$x=\frac{\sqrt{1330}}{4}+10\approx 19.117291264$$

$$x=-\frac{\sqrt{1330}}{4}+10\approx 0.882708736$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 32x by 20-x.

$$640x-32x^{2}=540$$

Subtract 540 from both sides.

$$640x-32x^{2}-540=0$$

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.

$$-32x^{2}+640x-540=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -32 for a, 640 for b, and -540 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-640±\sqrt{640^{2}-4\left(-32\right)\left(-540\right)}}{2\left(-32\right)}$$

Square 640.

$$x=\frac{-640±\sqrt{409600-4\left(-32\right)\left(-540\right)}}{2\left(-32\right)}$$

Multiply -4 times -32.

$$x=\frac{-640±\sqrt{409600+128\left(-540\right)}}{2\left(-32\right)}$$

Multiply 128 times -540.

$$x=\frac{-640±\sqrt{409600-69120}}{2\left(-32\right)}$$

Add 409600 to -69120.

$$x=\frac{-640±\sqrt{340480}}{2\left(-32\right)}$$

Take the square root of 340480.

$$x=\frac{-640±16\sqrt{1330}}{2\left(-32\right)}$$

Multiply 2 times -32.

$$x=\frac{-640±16\sqrt{1330}}{-64}$$

Now solve the equation x=\frac{-640±16\sqrt{1330}}{-64} when ± is plus. Add -640 to 16\sqrt{1330}.

$$x=\frac{16\sqrt{1330}-640}{-64}$$

Divide -640+16\sqrt{1330} by -64.

$$x=-\frac{\sqrt{1330}}{4}+10$$

Now solve the equation x=\frac{-640±16\sqrt{1330}}{-64} when ± is minus. Subtract 16\sqrt{1330} from -640.

$$x=\frac{-16\sqrt{1330}-640}{-64}$$

Divide -640-16\sqrt{1330} by -64.

$$x=\frac{\sqrt{1330}}{4}+10$$

The equation is now solved.

$$x=-\frac{\sqrt{1330}}{4}+10 x=\frac{\sqrt{1330}}{4}+10$$

Steps for Completing the Square

Use the distributive property to multiply 32x by 20-x.

$$640x-32x^{2}=540$$

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.

$$-32x^{2}+640x=540$$

Divide both sides by -32.

$$\frac{-32x^{2}+640x}{-32}=\frac{540}{-32}$$

Dividing by -32 undoes the multiplication by -32.

$$x^{2}+\frac{640}{-32}x=\frac{540}{-32}$$

Divide 640 by -32.

$$x^{2}-20x=\frac{540}{-32}$$

Reduce the fraction \frac{540}{-32} to lowest terms by extracting and canceling out 4.

$$x^{2}-20x=-\frac{135}{8}$$

Divide -20, the coefficient of the x term, by 2 to get -10. Then add the square of -10 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-20x+\left(-10\right)^{2}=-\frac{135}{8}+\left(-10\right)^{2}$$

Square -10.

$$x^{2}-20x+100=-\frac{135}{8}+100$$

Add -\frac{135}{8} to 100.

$$x^{2}-20x+100=\frac{665}{8}$$

Factor x^{2}-20x+100. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-10\right)^{2}=\frac{665}{8}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-10\right)^{2}}=\sqrt{\frac{665}{8}}$$

Simplify.

$$x-10=\frac{\sqrt{1330}}{4} x-10=-\frac{\sqrt{1330}}{4}$$

Add 10 to both sides of the equation.

$$x=\frac{\sqrt{1330}}{4}+10 x=-\frac{\sqrt{1330}}{4}+10$$