Solve 3=2n^2-n

$$3 = 2 { n }^{ 2 } -n$$

Solve for n

$$n=-1$$

$$n = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

Steps Using Factoring By Grouping

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

$$2n^{2}-n=3$$

Subtract 3 from both sides.

$$2n^{2}-n-3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2n^{2}+an+bn-3. To find a and b, set up a system to be solved.

$$a+b=-1 ab=2\left(-3\right)=-6$$

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 -6.

$$1,-6 2,-3$$

Calculate the sum for each pair.

$$1-6=-5 2-3=-1$$

The solution is the pair that gives sum -1.

$$a=-3 b=2$$

Rewrite 2n^{2}-n-3 as \left(2n^{2}-3n\right)+\left(2n-3\right).

$$\left(2n^{2}-3n\right)+\left(2n-3\right)$$

Factor out n in 2n^{2}-3n.

$$n\left(2n-3\right)+2n-3$$

Factor out common term 2n-3 by using distributive property.

$$\left(2n-3\right)\left(n+1\right)$$

To find equation solutions, solve 2n-3=0 and n+1=0.

$$n=\frac{3}{2} n=-1$$

Steps Using the Quadratic Formula

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

$$2n^{2}-n=3$$

Subtract 3 from both sides.

$$2n^{2}-n-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -1 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-\left(-1\right)±\sqrt{1-4\times 2\left(-3\right)}}{2\times 2}$$

Multiply -4 times 2.

$$n=\frac{-\left(-1\right)±\sqrt{1-8\left(-3\right)}}{2\times 2}$$

Multiply -8 times -3.

$$n=\frac{-\left(-1\right)±\sqrt{1+24}}{2\times 2}$$

Add 1 to 24.

$$n=\frac{-\left(-1\right)±\sqrt{25}}{2\times 2}$$

Take the square root of 25.

$$n=\frac{-\left(-1\right)±5}{2\times 2}$$

The opposite of -1 is 1.

$$n=\frac{1±5}{2\times 2}$$

Multiply 2 times 2.

$$n=\frac{1±5}{4}$$

Now solve the equation n=\frac{1±5}{4} when ± is plus. Add 1 to 5.

$$n=\frac{6}{4}$$

Reduce the fraction \frac{6}{4} to lowest terms by extracting and canceling out 2.

$$n=\frac{3}{2}$$

Now solve the equation n=\frac{1±5}{4} when ± is minus. Subtract 5 from 1.

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

Divide -4 by 4.

$$n=-1$$

The equation is now solved.

$$n=\frac{3}{2} n=-1$$

Steps for Completing the Square

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

$$2n^{2}-n=3$$

Divide both sides by 2.

$$\frac{2n^{2}-n}{2}=\frac{3}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$n^{2}-\frac{1}{2}n=\frac{3}{2}$$

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.

$$n^{2}-\frac{1}{2}n+\left(-\frac{1}{4}\right)^{2}=\frac{3}{2}+\left(-\frac{1}{4}\right)^{2}$$

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

$$n^{2}-\frac{1}{2}n+\frac{1}{16}=\frac{3}{2}+\frac{1}{16}$$

Add \frac{3}{2} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$n^{2}-\frac{1}{2}n+\frac{1}{16}=\frac{25}{16}$$

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}.

$$\left(n-\frac{1}{4}\right)^{2}=\frac{25}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n-\frac{1}{4}\right)^{2}}=\sqrt{\frac{25}{16}}$$

Simplify.

$$n-\frac{1}{4}=\frac{5}{4} n-\frac{1}{4}=-\frac{5}{4}$$

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

$$n=\frac{3}{2} n=-1$$