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

$$n = n ( 3 n - 1 ) / 2$$

Solve for n

$$n=1$$

$$n=0$$

Steps Using Factoring

Use the distributive property to multiply n by 3n-1.

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

Divide each term of 3n^{2}-n by 2 to get \frac{3}{2}n^{2}-\frac{1}{2}n.

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

Subtract \frac{3}{2}n^{2} from both sides.

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

Add \frac{1}{2}n to both sides.

$$n-\frac{3}{2}n^{2}+\frac{1}{2}n=0$$

Combine n and \frac{1}{2}n to get \frac{3}{2}n.

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

Factor out n.

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

To find equation solutions, solve n=0 and \frac{3-3n}{2}=0.

$$n=0 n=1$$

Steps Using the Quadratic Formula

Use the distributive property to multiply n by 3n-1.

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

Divide each term of 3n^{2}-n by 2 to get \frac{3}{2}n^{2}-\frac{1}{2}n.

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

Subtract \frac{3}{2}n^{2} from both sides.

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

Add \frac{1}{2}n to both sides.

$$n-\frac{3}{2}n^{2}+\frac{1}{2}n=0$$

Combine n and \frac{1}{2}n to get \frac{3}{2}n.

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

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.

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

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

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

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

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

Multiply 2 times -\frac{3}{2}.

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

Now solve the equation n=\frac{-\frac{3}{2}±\frac{3}{2}}{-3} when ± is plus. Add -\frac{3}{2} to \frac{3}{2} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

Divide 0 by -3.

$$n=0$$

Now solve the equation n=\frac{-\frac{3}{2}±\frac{3}{2}}{-3} when ± is minus. Subtract \frac{3}{2} from -\frac{3}{2} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

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

Divide -3 by -3.

$$n=1$$

The equation is now solved.

$$n=0 n=1$$

Steps for Completing the Square

Use the distributive property to multiply n by 3n-1.

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

Divide each term of 3n^{2}-n by 2 to get \frac{3}{2}n^{2}-\frac{1}{2}n.

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

Subtract \frac{3}{2}n^{2} from both sides.

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

Add \frac{1}{2}n to both sides.

$$n-\frac{3}{2}n^{2}+\frac{1}{2}n=0$$

Combine n and \frac{1}{2}n to get \frac{3}{2}n.

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

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.

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

Divide both sides of the equation by -\frac{3}{2}, which is the same as multiplying both sides by the reciprocal of the fraction.

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

Dividing by -\frac{3}{2} undoes the multiplication by -\frac{3}{2}.

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

Divide \frac{3}{2} by -\frac{3}{2} by multiplying \frac{3}{2} by the reciprocal of -\frac{3}{2}.

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

Divide 0 by -\frac{3}{2} by multiplying 0 by the reciprocal of -\frac{3}{2}.

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

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

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

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$n^{2}-n+\frac{1}{4}=\frac{1}{4}$$

Factor n^{2}-n+\frac{1}{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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add \frac{1}{2} to both sides of the equation.

$$n=1 n=0$$