Solve n(-344-4n+4)/2=(n-2)*180

$$\frac { n ( - 344 - 4 n + 4 ) } { 2 } = ( n - 2 ) \times 180$$

Solve for n

$$n = \frac{\sqrt{31345} - 175}{2} \approx 1.022595985$$

$$n=\frac{-\sqrt{31345}-175}{2}\approx -176.022595985$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 2.

$$n\left(-344-4n+4\right)=2\left(n-2\right)\times 180$$

Add -344 and 4 to get -340.

$$n\left(-340-4n\right)=2\left(n-2\right)\times 180$$

Use the distributive property to multiply n by -340-4n.

$$-340n-4n^{2}=2\left(n-2\right)\times 180$$

Multiply 2 and 180 to get 360.

$$-340n-4n^{2}=360\left(n-2\right)$$

Use the distributive property to multiply 360 by n-2.

$$-340n-4n^{2}=360n-720$$

Subtract 360n from both sides.

$$-340n-4n^{2}-360n=-720$$

Combine -340n and -360n to get -700n.

$$-700n-4n^{2}=-720$$

Add 720 to both sides.

$$-700n-4n^{2}+720=0$$

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.

$$-4n^{2}-700n+720=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, -700 for b, and 720 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-\left(-700\right)±\sqrt{\left(-700\right)^{2}-4\left(-4\right)\times 720}}{2\left(-4\right)}$$

Square -700.

$$n=\frac{-\left(-700\right)±\sqrt{490000-4\left(-4\right)\times 720}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$n=\frac{-\left(-700\right)±\sqrt{490000+16\times 720}}{2\left(-4\right)}$$

Multiply 16 times 720.

$$n=\frac{-\left(-700\right)±\sqrt{490000+11520}}{2\left(-4\right)}$$

Add 490000 to 11520.

$$n=\frac{-\left(-700\right)±\sqrt{501520}}{2\left(-4\right)}$$

Take the square root of 501520.

$$n=\frac{-\left(-700\right)±4\sqrt{31345}}{2\left(-4\right)}$$

The opposite of -700 is 700.

$$n=\frac{700±4\sqrt{31345}}{2\left(-4\right)}$$

Multiply 2 times -4.

$$n=\frac{700±4\sqrt{31345}}{-8}$$

Now solve the equation n=\frac{700±4\sqrt{31345}}{-8} when ± is plus. Add 700 to 4\sqrt{31345}.

$$n=\frac{4\sqrt{31345}+700}{-8}$$

Divide 700+4\sqrt{31345} by -8.

$$n=\frac{-\sqrt{31345}-175}{2}$$

Now solve the equation n=\frac{700±4\sqrt{31345}}{-8} when ± is minus. Subtract 4\sqrt{31345} from 700.

$$n=\frac{700-4\sqrt{31345}}{-8}$$

Divide 700-4\sqrt{31345} by -8.

$$n=\frac{\sqrt{31345}-175}{2}$$

The equation is now solved.

$$n=\frac{-\sqrt{31345}-175}{2} n=\frac{\sqrt{31345}-175}{2}$$

Steps for Completing the Square

Multiply both sides of the equation by 2.

$$n\left(-344-4n+4\right)=2\left(n-2\right)\times 180$$

Add -344 and 4 to get -340.

$$n\left(-340-4n\right)=2\left(n-2\right)\times 180$$

Use the distributive property to multiply n by -340-4n.

$$-340n-4n^{2}=2\left(n-2\right)\times 180$$

Multiply 2 and 180 to get 360.

$$-340n-4n^{2}=360\left(n-2\right)$$

Use the distributive property to multiply 360 by n-2.

$$-340n-4n^{2}=360n-720$$

Subtract 360n from both sides.

$$-340n-4n^{2}-360n=-720$$

Combine -340n and -360n to get -700n.

$$-700n-4n^{2}=-720$$

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.

$$-4n^{2}-700n=-720$$

Divide both sides by -4.

$$\frac{-4n^{2}-700n}{-4}=-\frac{720}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$n^{2}+\left(-\frac{700}{-4}\right)n=-\frac{720}{-4}$$

Divide -700 by -4.

$$n^{2}+175n=-\frac{720}{-4}$$

Divide -720 by -4.

$$n^{2}+175n=180$$

Divide 175, the coefficient of the x term, by 2 to get \frac{175}{2}. Then add the square of \frac{175}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}+175n+\left(\frac{175}{2}\right)^{2}=180+\left(\frac{175}{2}\right)^{2}$$

Square \frac{175}{2} by squaring both the numerator and the denominator of the fraction.

$$n^{2}+175n+\frac{30625}{4}=180+\frac{30625}{4}$$

Add 180 to \frac{30625}{4}.

$$n^{2}+175n+\frac{30625}{4}=\frac{31345}{4}$$

Factor n^{2}+175n+\frac{30625}{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}.

$$\left(n+\frac{175}{2}\right)^{2}=\frac{31345}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n+\frac{175}{2}\right)^{2}}=\sqrt{\frac{31345}{4}}$$

Simplify.

$$n+\frac{175}{2}=\frac{\sqrt{31345}}{2} n+\frac{175}{2}=-\frac{\sqrt{31345}}{2}$$

Subtract \frac{175}{2} from both sides of the equation.

$$n=\frac{\sqrt{31345}-175}{2} n=\frac{-\sqrt{31345}-175}{2}$$