Solve 2(n^2+n)=5n

$$2 ( n ^ { 2 } + n ) = 5 n$$

Solve for n

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

$$n=0$$

Steps Using Factoring

Use the distributive property to multiply 2 by n^{2}+n.

$$2n^{2}+2n=5n$$

Subtract 5n from both sides.

$$2n^{2}+2n-5n=0$$

Combine 2n and -5n to get -3n.

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

Factor out n.

$$n\left(2n-3\right)=0$$

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

$$n=0 n=\frac{3}{2}$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 2 by n^{2}+n.

$$2n^{2}+2n=5n$$

Subtract 5n from both sides.

$$2n^{2}+2n-5n=0$$

Combine 2n and -5n to get -3n.

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

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

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

Take the square root of \left(-3\right)^{2}.

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

The opposite of -3 is 3.

$$n=\frac{3±3}{2\times 2}$$

Multiply 2 times 2.

$$n=\frac{3±3}{4}$$

Now solve the equation n=\frac{3±3}{4} when ± is plus. Add 3 to 3.

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

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{3±3}{4} when ± is minus. Subtract 3 from 3.

$$n=\frac{0}{4}$$

Divide 0 by 4.

$$n=0$$

The equation is now solved.

$$n=\frac{3}{2} n=0$$

Steps for Completing the Square

Use the distributive property to multiply 2 by n^{2}+n.

$$2n^{2}+2n=5n$$

Subtract 5n from both sides.

$$2n^{2}+2n-5n=0$$

Combine 2n and -5n to get -3n.

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide 0 by 2.

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

Divide -\frac{3}{2}, the coefficient of the x term, by 2 to get -\frac{3}{4}. Then add the square of -\frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$n^{2}-\frac{3}{2}n+\frac{9}{16}=\frac{9}{16}$$

Factor n^{2}-\frac{3}{2}n+\frac{9}{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}.

$$\left(n-\frac{3}{4}\right)^{2}=\frac{9}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n-\frac{3}{4}\right)^{2}}=\sqrt{\frac{9}{16}}$$

Simplify.

$$n-\frac{3}{4}=\frac{3}{4} n-\frac{3}{4}=-\frac{3}{4}$$

Add \frac{3}{4} to both sides of the equation.

$$n=\frac{3}{2} n=0$$