Solve n(n-3)/2=n/1Rightarrown=?

$$\frac { n ( n - 3 ) } { 2 } = \frac { n } { 1 } \Rightarrow n = ?$$

Solve for n

$$n=5$$

$$n=0$$

Steps Using Factoring

Multiply both sides of the equation by 2.

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

Use the distributive property to multiply n by n-3.

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

Subtract 2n from both sides.

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

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

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

Factor out n.

$$n\left(n-5\right)=0$$

To find equation solutions, solve n=0 and n-5=0.

$$n=0 n=5$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 2.

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

Use the distributive property to multiply n by n-3.

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

Subtract 2n from both sides.

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

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

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -5 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}}}{2}$$

Take the square root of \left(-5\right)^{2}.

$$n=\frac{-\left(-5\right)±5}{2}$$

The opposite of -5 is 5.

$$n=\frac{5±5}{2}$$

Now solve the equation n=\frac{5±5}{2} when ± is plus. Add 5 to 5.

$$n=\frac{10}{2}$$

Divide 10 by 2.

$$n=5$$

Now solve the equation n=\frac{5±5}{2} when ± is minus. Subtract 5 from 5.

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

Divide 0 by 2.

$$n=0$$

The equation is now solved.

$$n=5 n=0$$

Steps for Completing the Square

Multiply both sides of the equation by 2.

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

Use the distributive property to multiply n by n-3.

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

Subtract 2n from both sides.

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

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

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

Divide -5, the coefficient of the x term, by 2 to get -\frac{5}{2}. Then add the square of -\frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}-5n+\left(-\frac{5}{2}\right)^{2}=\left(-\frac{5}{2}\right)^{2}$$

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$n^{2}-5n+\frac{25}{4}=\frac{25}{4}$$

Factor n^{2}-5n+\frac{25}{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}.

$$\left(n-\frac{5}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n-\frac{5}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$n-\frac{5}{2}=\frac{5}{2} n-\frac{5}{2}=-\frac{5}{2}$$

Add \frac{5}{2} to both sides of the equation.

$$n=5 n=0$$