Solve (18-x)(320-10x)=400

$$(18-x)(320-10x)=400$$

Solve for x

$$x=\sqrt{89}+25\approx 34.433981132$$

$$x=25-\sqrt{89}\approx 15.566018868$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 18-x by 320-10x and combine like terms.

$$5760-500x+10x^{2}=400$$

Subtract 400 from both sides.

$$5760-500x+10x^{2}-400=0$$

Subtract 400 from 5760 to get 5360.

$$5360-500x+10x^{2}=0$$

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.

$$10x^{2}-500x+5360=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 10 for a, -500 for b, and 5360 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-500\right)±\sqrt{\left(-500\right)^{2}-4\times 10\times 5360}}{2\times 10}$$

Square -500.

$$x=\frac{-\left(-500\right)±\sqrt{250000-4\times 10\times 5360}}{2\times 10}$$

Multiply -4 times 10.

$$x=\frac{-\left(-500\right)±\sqrt{250000-40\times 5360}}{2\times 10}$$

Multiply -40 times 5360.

$$x=\frac{-\left(-500\right)±\sqrt{250000-214400}}{2\times 10}$$

Add 250000 to -214400.

$$x=\frac{-\left(-500\right)±\sqrt{35600}}{2\times 10}$$

Take the square root of 35600.

$$x=\frac{-\left(-500\right)±20\sqrt{89}}{2\times 10}$$

The opposite of -500 is 500.

$$x=\frac{500±20\sqrt{89}}{2\times 10}$$

Multiply 2 times 10.

$$x=\frac{500±20\sqrt{89}}{20}$$

Now solve the equation x=\frac{500±20\sqrt{89}}{20} when ± is plus. Add 500 to 20\sqrt{89}.

$$x=\frac{20\sqrt{89}+500}{20}$$

Divide 500+20\sqrt{89} by 20.

$$x=\sqrt{89}+25$$

Now solve the equation x=\frac{500±20\sqrt{89}}{20} when ± is minus. Subtract 20\sqrt{89} from 500.

$$x=\frac{500-20\sqrt{89}}{20}$$

Divide 500-20\sqrt{89} by 20.

$$x=25-\sqrt{89}$$

The equation is now solved.

$$x=\sqrt{89}+25 x=25-\sqrt{89}$$

Steps for Completing the Square

Use the distributive property to multiply 18-x by 320-10x and combine like terms.

$$5760-500x+10x^{2}=400$$

Subtract 5760 from both sides.

$$-500x+10x^{2}=400-5760$$

Subtract 5760 from 400 to get -5360.

$$-500x+10x^{2}=-5360$$

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.

$$10x^{2}-500x=-5360$$

Divide both sides by 10.

$$\frac{10x^{2}-500x}{10}=-\frac{5360}{10}$$

Dividing by 10 undoes the multiplication by 10.

$$x^{2}+\left(-\frac{500}{10}\right)x=-\frac{5360}{10}$$

Divide -500 by 10.

$$x^{2}-50x=-\frac{5360}{10}$$

Divide -5360 by 10.

$$x^{2}-50x=-536$$

Divide -50, the coefficient of the x term, by 2 to get -25. Then add the square of -25 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-50x+\left(-25\right)^{2}=-536+\left(-25\right)^{2}$$

Square -25.

$$x^{2}-50x+625=-536+625$$

Add -536 to 625.

$$x^{2}-50x+625=89$$

Factor x^{2}-50x+625. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-25\right)^{2}=89$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-25\right)^{2}}=\sqrt{89}$$

Simplify.

$$x-25=\sqrt{89} x-25=-\sqrt{89}$$

Add 25 to both sides of the equation.

$$x=\sqrt{89}+25 x=25-\sqrt{89}$$