Solve (3-2+x)(500-100x)=800

$$(3-2+x)(500-100x)=800$$

Solve for x

$$x=1$$

$$x=3$$

Steps Using the Quadratic Formula

Subtract 2 from 3 to get 1.

$$\left(1+x\right)\left(500-100x\right)=800$$

Use the distributive property to multiply 1+x by 500-100x and combine like terms.

$$500+400x-100x^{2}=800$$

Subtract 800 from both sides.

$$500+400x-100x^{2}-800=0$$

Subtract 800 from 500 to get -300.

$$-300+400x-100x^{2}=0$$

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.

$$-100x^{2}+400x-300=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -100 for a, 400 for b, and -300 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-400±\sqrt{400^{2}-4\left(-100\right)\left(-300\right)}}{2\left(-100\right)}$$

Square 400.

$$x=\frac{-400±\sqrt{160000-4\left(-100\right)\left(-300\right)}}{2\left(-100\right)}$$

Multiply -4 times -100.

$$x=\frac{-400±\sqrt{160000+400\left(-300\right)}}{2\left(-100\right)}$$

Multiply 400 times -300.

$$x=\frac{-400±\sqrt{160000-120000}}{2\left(-100\right)}$$

Add 160000 to -120000.

$$x=\frac{-400±\sqrt{40000}}{2\left(-100\right)}$$

Take the square root of 40000.

$$x=\frac{-400±200}{2\left(-100\right)}$$

Multiply 2 times -100.

$$x=\frac{-400±200}{-200}$$

Now solve the equation x=\frac{-400±200}{-200} when ± is plus. Add -400 to 200.

$$x=-\frac{200}{-200}$$

Divide -200 by -200.

$$x=1$$

Now solve the equation x=\frac{-400±200}{-200} when ± is minus. Subtract 200 from -400.

$$x=-\frac{600}{-200}$$

Divide -600 by -200.

$$x=3$$

The equation is now solved.

$$x=1 x=3$$

Steps for Completing the Square

Subtract 2 from 3 to get 1.

$$\left(1+x\right)\left(500-100x\right)=800$$

Use the distributive property to multiply 1+x by 500-100x and combine like terms.

$$500+400x-100x^{2}=800$$

Subtract 500 from both sides.

$$400x-100x^{2}=800-500$$

Subtract 500 from 800 to get 300.

$$400x-100x^{2}=300$$

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.

$$-100x^{2}+400x=300$$

Divide both sides by -100.

$$\frac{-100x^{2}+400x}{-100}=\frac{300}{-100}$$

Dividing by -100 undoes the multiplication by -100.

$$x^{2}+\frac{400}{-100}x=\frac{300}{-100}$$

Divide 400 by -100.

$$x^{2}-4x=\frac{300}{-100}$$

Divide 300 by -100.

$$x^{2}-4x=-3$$

Divide -4, the coefficient of the x term, by 2 to get -2. Then add the square of -2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-4x+\left(-2\right)^{2}=-3+\left(-2\right)^{2}$$

Square -2.

$$x^{2}-4x+4=-3+4$$

Add -3 to 4.

$$x^{2}-4x+4=1$$

Factor x^{2}-4x+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-2\right)^{2}=1$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{1}$$

Simplify.

$$x-2=1 x-2=-1$$

Add 2 to both sides of the equation.

$$x=3 x=1$$