Solve left(500-left(x-40right)*10right)left(x-40right)=8000

$$\left( 500- \left( x-40 \right) \times 10 \right) \left( x-40 \right) = 8000$$

Solve for x (complex solution)

$$x=65+5\sqrt{7}i\approx 65+13.228756555i$$

$$x=-5\sqrt{7}i+65\approx 65-13.228756555i$$

Steps Using the Quadratic Formula

Use the distributive property to multiply x-40 by 10.

$$\left(500-\left(10x-400\right)\right)\left(x-40\right)=8000$$

To find the opposite of 10x-400, find the opposite of each term.

$$\left(500-10x-\left(-400\right)\right)\left(x-40\right)=8000$$

The opposite of -400 is 400.

$$\left(500-10x+400\right)\left(x-40\right)=8000$$

Add 500 and 400 to get 900.

$$\left(900-10x\right)\left(x-40\right)=8000$$

Apply the distributive property by multiplying each term of 900-10x by each term of x-40.

$$900x-36000-10x^{2}+400x=8000$$

Combine 900x and 400x to get 1300x.

$$1300x-36000-10x^{2}=8000$$

Subtract 8000 from both sides.

$$1300x-36000-10x^{2}-8000=0$$

Subtract 8000 from -36000 to get -44000.

$$1300x-44000-10x^{2}=0$$

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.

$$-10x^{2}+1300x-44000=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -10 for a, 1300 for b, and -44000 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1300±\sqrt{1300^{2}-4\left(-10\right)\left(-44000\right)}}{2\left(-10\right)}$$

Square 1300.

$$x=\frac{-1300±\sqrt{1690000-4\left(-10\right)\left(-44000\right)}}{2\left(-10\right)}$$

Multiply -4 times -10.

$$x=\frac{-1300±\sqrt{1690000+40\left(-44000\right)}}{2\left(-10\right)}$$

Multiply 40 times -44000.

$$x=\frac{-1300±\sqrt{1690000-1760000}}{2\left(-10\right)}$$

Add 1690000 to -1760000.

$$x=\frac{-1300±\sqrt{-70000}}{2\left(-10\right)}$$

Take the square root of -70000.

$$x=\frac{-1300±100\sqrt{7}i}{2\left(-10\right)}$$

Multiply 2 times -10.

$$x=\frac{-1300±100\sqrt{7}i}{-20}$$

Now solve the equation x=\frac{-1300±100\sqrt{7}i}{-20} when ± is plus. Add -1300 to 100i\sqrt{7}.

$$x=\frac{-1300+100\sqrt{7}i}{-20}$$

Divide -1300+100i\sqrt{7} by -20.

$$x=-5\sqrt{7}i+65$$

Now solve the equation x=\frac{-1300±100\sqrt{7}i}{-20} when ± is minus. Subtract 100i\sqrt{7} from -1300.

$$x=\frac{-100\sqrt{7}i-1300}{-20}$$

Divide -1300-100i\sqrt{7} by -20.

$$x=65+5\sqrt{7}i$$

The equation is now solved.

$$x=-5\sqrt{7}i+65 x=65+5\sqrt{7}i$$

Steps for Completing the Square

Use the distributive property to multiply x-40 by 10.

$$\left(500-\left(10x-400\right)\right)\left(x-40\right)=8000$$

To find the opposite of 10x-400, find the opposite of each term.

$$\left(500-10x-\left(-400\right)\right)\left(x-40\right)=8000$$

The opposite of -400 is 400.

$$\left(500-10x+400\right)\left(x-40\right)=8000$$

Add 500 and 400 to get 900.

$$\left(900-10x\right)\left(x-40\right)=8000$$

Apply the distributive property by multiplying each term of 900-10x by each term of x-40.

$$900x-36000-10x^{2}+400x=8000$$

Combine 900x and 400x to get 1300x.

$$1300x-36000-10x^{2}=8000$$

Add 36000 to both sides.

$$1300x-10x^{2}=8000+36000$$

Add 8000 and 36000 to get 44000.

$$1300x-10x^{2}=44000$$

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.

$$-10x^{2}+1300x=44000$$

Divide both sides by -10.

$$\frac{-10x^{2}+1300x}{-10}=\frac{44000}{-10}$$

Dividing by -10 undoes the multiplication by -10.

$$x^{2}+\frac{1300}{-10}x=\frac{44000}{-10}$$

Divide 1300 by -10.

$$x^{2}-130x=\frac{44000}{-10}$$

Divide 44000 by -10.

$$x^{2}-130x=-4400$$

Divide -130, the coefficient of the x term, by 2 to get -65. Then add the square of -65 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-130x+\left(-65\right)^{2}=-4400+\left(-65\right)^{2}$$

Square -65.

$$x^{2}-130x+4225=-4400+4225$$

Add -4400 to 4225.

$$x^{2}-130x+4225=-175$$

Factor x^{2}-130x+4225. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-65\right)^{2}=-175$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-65\right)^{2}}=\sqrt{-175}$$

Simplify.

$$x-65=5\sqrt{7}i x-65=-5\sqrt{7}i$$

Add 65 to both sides of the equation.

$$x=65+5\sqrt{7}i x=-5\sqrt{7}i+65$$