Solve n(n-1)/2=54+n

$$\frac { n ( n - 1 ) } { 2 } = 54 + n$$

Solve for n

$$n=-9$$

$$n=12$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 2.

$$n\left(n-1\right)=108+2n$$

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

$$n^{2}-n=108+2n$$

Subtract 108 from both sides.

$$n^{2}-n-108=2n$$

Subtract 2n from both sides.

$$n^{2}-n-108-2n=0$$

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

$$n^{2}-3n-108=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -3 for b, and -108 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-108\right)}}{2}$$

Square -3.

$$n=\frac{-\left(-3\right)±\sqrt{9-4\left(-108\right)}}{2}$$

Multiply -4 times -108.

$$n=\frac{-\left(-3\right)±\sqrt{9+432}}{2}$$

Add 9 to 432.

$$n=\frac{-\left(-3\right)±\sqrt{441}}{2}$$

Take the square root of 441.

$$n=\frac{-\left(-3\right)±21}{2}$$

The opposite of -3 is 3.

$$n=\frac{3±21}{2}$$

Now solve the equation n=\frac{3±21}{2} when ± is plus. Add 3 to 21.

$$n=\frac{24}{2}$$

Divide 24 by 2.

$$n=12$$

Now solve the equation n=\frac{3±21}{2} when ± is minus. Subtract 21 from 3.

$$n=-\frac{18}{2}$$

Divide -18 by 2.

$$n=-9$$

The equation is now solved.

$$n=12 n=-9$$

Steps for Completing the Square

Multiply both sides of the equation by 2.

$$n\left(n-1\right)=108+2n$$

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

$$n^{2}-n=108+2n$$

Subtract 2n from both sides.

$$n^{2}-n-2n=108$$

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

$$n^{2}-3n=108$$

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

$$n^{2}-3n+\left(-\frac{3}{2}\right)^{2}=108+\left(-\frac{3}{2}\right)^{2}$$

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$n^{2}-3n+\frac{9}{4}=108+\frac{9}{4}$$

Add 108 to \frac{9}{4}.

$$n^{2}-3n+\frac{9}{4}=\frac{441}{4}$$

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

$$\left(n-\frac{3}{2}\right)^{2}=\frac{441}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n-\frac{3}{2}\right)^{2}}=\sqrt{\frac{441}{4}}$$

Simplify.

$$n-\frac{3}{2}=\frac{21}{2} n-\frac{3}{2}=-\frac{21}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$n=12 n=-9$$