Solve 1/2n(1000+(n-1)*200)=32000

$$\frac { 1 } { 2 } n ( 1000 + ( n - 1 ) \times 200 ) = 32000$$

Solve for n

$$n=-20$$

$$n=16$$

Steps Using the Quadratic Formula

Use the distributive property to multiply n-1 by 200.

$$\frac{1}{2}n\left(1000+200n-200\right)=32000$$

Subtract 200 from 1000 to get 800.

$$\frac{1}{2}n\left(800+200n\right)=32000$$

Use the distributive property to multiply \frac{1}{2}n by 800+200n.

$$\frac{1}{2}n\times 800+\frac{1}{2}n\times 200n=32000$$

Multiply n and n to get n^{2}.

$$\frac{1}{2}n\times 800+\frac{1}{2}n^{2}\times 200=32000$$

Multiply \frac{1}{2} and 800 to get \frac{800}{2}.

$$\frac{800}{2}n+\frac{1}{2}n^{2}\times 200=32000$$

Divide 800 by 2 to get 400.

$$400n+\frac{1}{2}n^{2}\times 200=32000$$

Multiply \frac{1}{2} and 200 to get \frac{200}{2}.

$$400n+\frac{200}{2}n^{2}=32000$$

Divide 200 by 2 to get 100.

$$400n+100n^{2}=32000$$

Subtract 32000 from both sides.

$$400n+100n^{2}-32000=0$$

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.

$$100n^{2}+400n-32000=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 100 for a, 400 for b, and -32000 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-400±\sqrt{400^{2}-4\times 100\left(-32000\right)}}{2\times 100}$$

Square 400.

$$n=\frac{-400±\sqrt{160000-4\times 100\left(-32000\right)}}{2\times 100}$$

Multiply -4 times 100.

$$n=\frac{-400±\sqrt{160000-400\left(-32000\right)}}{2\times 100}$$

Multiply -400 times -32000.

$$n=\frac{-400±\sqrt{160000+12800000}}{2\times 100}$$

Add 160000 to 12800000.

$$n=\frac{-400±\sqrt{12960000}}{2\times 100}$$

Take the square root of 12960000.

$$n=\frac{-400±3600}{2\times 100}$$

Multiply 2 times 100.

$$n=\frac{-400±3600}{200}$$

Now solve the equation n=\frac{-400±3600}{200} when ± is plus. Add -400 to 3600.

$$n=\frac{3200}{200}$$

Divide 3200 by 200.

$$n=16$$

Now solve the equation n=\frac{-400±3600}{200} when ± is minus. Subtract 3600 from -400.

$$n=-\frac{4000}{200}$$

Divide -4000 by 200.

$$n=-20$$

The equation is now solved.

$$n=16 n=-20$$

Steps for Completing the Square

Use the distributive property to multiply n-1 by 200.

$$\frac{1}{2}n\left(1000+200n-200\right)=32000$$

Subtract 200 from 1000 to get 800.

$$\frac{1}{2}n\left(800+200n\right)=32000$$

Use the distributive property to multiply \frac{1}{2}n by 800+200n.

$$\frac{1}{2}n\times 800+\frac{1}{2}n\times 200n=32000$$

Multiply n and n to get n^{2}.

$$\frac{1}{2}n\times 800+\frac{1}{2}n^{2}\times 200=32000$$

Multiply \frac{1}{2} and 800 to get \frac{800}{2}.

$$\frac{800}{2}n+\frac{1}{2}n^{2}\times 200=32000$$

Divide 800 by 2 to get 400.

$$400n+\frac{1}{2}n^{2}\times 200=32000$$

Multiply \frac{1}{2} and 200 to get \frac{200}{2}.

$$400n+\frac{200}{2}n^{2}=32000$$

Divide 200 by 2 to get 100.

$$400n+100n^{2}=32000$$

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.

$$100n^{2}+400n=32000$$

Divide both sides by 100.

$$\frac{100n^{2}+400n}{100}=\frac{32000}{100}$$

Dividing by 100 undoes the multiplication by 100.

$$n^{2}+\frac{400}{100}n=\frac{32000}{100}$$

Divide 400 by 100.

$$n^{2}+4n=\frac{32000}{100}$$

Divide 32000 by 100.

$$n^{2}+4n=320$$

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.

$$n^{2}+4n+2^{2}=320+2^{2}$$

Square 2.

$$n^{2}+4n+4=320+4$$

Add 320 to 4.

$$n^{2}+4n+4=324$$

Factor n^{2}+4n+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}.

$$\left(n+2\right)^{2}=324$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n+2\right)^{2}}=\sqrt{324}$$

Simplify.

$$n+2=18 n+2=-18$$

Subtract 2 from both sides of the equation.

$$n=16 n=-20$$