Solve n(84+(n-1)-8)/2=-140

$$\frac { n ( 84 + ( n - 1 ) - 8 ) } { 2 } = - 140$$

Solve for n

$$n=\frac{\sqrt{4505}-75}{2}\approx -3.940351611$$

$$n=\frac{-\sqrt{4505}-75}{2}\approx -71.059648389$$

Steps Using the Quadratic Formula

Multiply both sides by 2.

$$n\left(84+n-1-8\right)=-140\times 2$$

Subtract 1 from 84 to get 83.

$$n\left(83+n-8\right)=-140\times 2$$

Subtract 8 from 83 to get 75.

$$n\left(75+n\right)=-140\times 2$$

Use the distributive property to multiply n by 75+n.

$$75n+n^{2}=-140\times 2$$

Multiply -140 and 2 to get -280.

$$75n+n^{2}=-280$$

Add 280 to both sides.

$$75n+n^{2}+280=0$$

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.

$$n^{2}+75n+280=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 75 for b, and 280 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-75±\sqrt{75^{2}-4\times 280}}{2}$$

Square 75.

$$n=\frac{-75±\sqrt{5625-4\times 280}}{2}$$

Multiply -4 times 280.

$$n=\frac{-75±\sqrt{5625-1120}}{2}$$

Add 5625 to -1120.

$$n=\frac{-75±\sqrt{4505}}{2}$$

Now solve the equation n=\frac{-75±\sqrt{4505}}{2} when ± is plus. Add -75 to \sqrt{4505}.

$$n=\frac{\sqrt{4505}-75}{2}$$

Now solve the equation n=\frac{-75±\sqrt{4505}}{2} when ± is minus. Subtract \sqrt{4505} from -75.

$$n=\frac{-\sqrt{4505}-75}{2}$$

The equation is now solved.

$$n=\frac{\sqrt{4505}-75}{2} n=\frac{-\sqrt{4505}-75}{2}$$

Steps for Completing the Square

Multiply both sides by 2.

$$n\left(84+n-1-8\right)=-140\times 2$$

Subtract 1 from 84 to get 83.

$$n\left(83+n-8\right)=-140\times 2$$

Subtract 8 from 83 to get 75.

$$n\left(75+n\right)=-140\times 2$$

Use the distributive property to multiply n by 75+n.

$$75n+n^{2}=-140\times 2$$

Multiply -140 and 2 to get -280.

$$75n+n^{2}=-280$$

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.

$$n^{2}+75n=-280$$

Divide 75, the coefficient of the x term, by 2 to get \frac{75}{2}. Then add the square of \frac{75}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}+75n+\left(\frac{75}{2}\right)^{2}=-280+\left(\frac{75}{2}\right)^{2}$$

Square \frac{75}{2} by squaring both the numerator and the denominator of the fraction.

$$n^{2}+75n+\frac{5625}{4}=-280+\frac{5625}{4}$$

Add -280 to \frac{5625}{4}.

$$n^{2}+75n+\frac{5625}{4}=\frac{4505}{4}$$

Factor n^{2}+75n+\frac{5625}{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}.

$$\left(n+\frac{75}{2}\right)^{2}=\frac{4505}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n+\frac{75}{2}\right)^{2}}=\sqrt{\frac{4505}{4}}$$

Simplify.

$$n+\frac{75}{2}=\frac{\sqrt{4505}}{2} n+\frac{75}{2}=-\frac{\sqrt{4505}}{2}$$

Subtract \frac{75}{2} from both sides of the equation.

$$n=\frac{\sqrt{4505}-75}{2} n=\frac{-\sqrt{4505}-75}{2}$$