Solve 561=2n^2-n

$$561 = 2 n ^ { 2 } - n$$

Solve for n

$$n = -\frac{33}{2} = -16\frac{1}{2} = -16.5$$

$$n=17$$

Steps Using Factoring By Grouping

Swap sides so that all variable terms are on the left hand side.

$$2n^{2}-n=561$$

Subtract 561 from both sides.

$$2n^{2}-n-561=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2n^{2}+an+bn-561. To find a and b, set up a system to be solved.

$$a+b=-1 ab=2\left(-561\right)=-1122$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -1122.

$$1,-1122 2,-561 3,-374 6,-187 11,-102 17,-66 22,-51 33,-34$$

Calculate the sum for each pair.

$$1-1122=-1121 2-561=-559 3-374=-371 6-187=-181 11-102=-91 17-66=-49 22-51=-29 33-34=-1$$

The solution is the pair that gives sum -1.

$$a=-34 b=33$$

Rewrite 2n^{2}-n-561 as \left(2n^{2}-34n\right)+\left(33n-561\right).

$$\left(2n^{2}-34n\right)+\left(33n-561\right)$$

Factor out 2n in the first and 33 in the second group.

$$2n\left(n-17\right)+33\left(n-17\right)$$

Factor out common term n-17 by using distributive property.

$$\left(n-17\right)\left(2n+33\right)$$

To find equation solutions, solve n-17=0 and 2n+33=0.

$$n=17 n=-\frac{33}{2}$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$2n^{2}-n=561$$

Subtract 561 from both sides.

$$2n^{2}-n-561=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -1 for b, and -561 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-\left(-1\right)±\sqrt{1-4\times 2\left(-561\right)}}{2\times 2}$$

Multiply -4 times 2.

$$n=\frac{-\left(-1\right)±\sqrt{1-8\left(-561\right)}}{2\times 2}$$

Multiply -8 times -561.

$$n=\frac{-\left(-1\right)±\sqrt{1+4488}}{2\times 2}$$

Add 1 to 4488.

$$n=\frac{-\left(-1\right)±\sqrt{4489}}{2\times 2}$$

Take the square root of 4489.

$$n=\frac{-\left(-1\right)±67}{2\times 2}$$

The opposite of -1 is 1.

$$n=\frac{1±67}{2\times 2}$$

Multiply 2 times 2.

$$n=\frac{1±67}{4}$$

Now solve the equation n=\frac{1±67}{4} when ± is plus. Add 1 to 67.

$$n=\frac{68}{4}$$

Divide 68 by 4.

$$n=17$$

Now solve the equation n=\frac{1±67}{4} when ± is minus. Subtract 67 from 1.

$$n=-\frac{66}{4}$$

Reduce the fraction \frac{-66}{4} to lowest terms by extracting and canceling out 2.

$$n=-\frac{33}{2}$$

The equation is now solved.

$$n=17 n=-\frac{33}{2}$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

$$2n^{2}-n=561$$

Divide both sides by 2.

$$\frac{2n^{2}-n}{2}=\frac{561}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$n^{2}-\frac{1}{2}n=\frac{561}{2}$$

Divide -\frac{1}{2}, the coefficient of the x term, by 2 to get -\frac{1}{4}. Then add the square of -\frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}-\frac{1}{2}n+\left(-\frac{1}{4}\right)^{2}=\frac{561}{2}+\left(-\frac{1}{4}\right)^{2}$$

Square -\frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$n^{2}-\frac{1}{2}n+\frac{1}{16}=\frac{561}{2}+\frac{1}{16}$$

Add \frac{561}{2} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$n^{2}-\frac{1}{2}n+\frac{1}{16}=\frac{4489}{16}$$

Factor n^{2}-\frac{1}{2}n+\frac{1}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(n-\frac{1}{4}\right)^{2}=\frac{4489}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n-\frac{1}{4}\right)^{2}}=\sqrt{\frac{4489}{16}}$$

Simplify.

$$n-\frac{1}{4}=\frac{67}{4} n-\frac{1}{4}=-\frac{67}{4}$$

Add \frac{1}{4} to both sides of the equation.

$$n=17 n=-\frac{33}{2}$$