Solve 2n+n(n-1)/2*3=2015

$$2 n + \frac { n ( n - 1 ) } { 2 } \times 3 = 2015$$

Solve for n

$$n = \frac{\sqrt{48361} - 1}{6} \approx 36.485224296$$

$$n=\frac{-\sqrt{48361}-1}{6}\approx -36.818557629$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 2.

$$4n+n\left(n-1\right)\times 3=4030$$

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

$$4n+\left(n^{2}-n\right)\times 3=4030$$

Use the distributive property to multiply n^{2}-n by 3.

$$4n+3n^{2}-3n=4030$$

Combine 4n and -3n to get n.

$$n+3n^{2}=4030$$

Subtract 4030 from both sides.

$$n+3n^{2}-4030=0$$

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.

$$3n^{2}+n-4030=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 1 for b, and -4030 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-1±\sqrt{1^{2}-4\times 3\left(-4030\right)}}{2\times 3}$$

Square 1.

$$n=\frac{-1±\sqrt{1-4\times 3\left(-4030\right)}}{2\times 3}$$

Multiply -4 times 3.

$$n=\frac{-1±\sqrt{1-12\left(-4030\right)}}{2\times 3}$$

Multiply -12 times -4030.

$$n=\frac{-1±\sqrt{1+48360}}{2\times 3}$$

Add 1 to 48360.

$$n=\frac{-1±\sqrt{48361}}{2\times 3}$$

Multiply 2 times 3.

$$n=\frac{-1±\sqrt{48361}}{6}$$

Now solve the equation n=\frac{-1±\sqrt{48361}}{6} when ± is plus. Add -1 to \sqrt{48361}.

$$n=\frac{\sqrt{48361}-1}{6}$$

Now solve the equation n=\frac{-1±\sqrt{48361}}{6} when ± is minus. Subtract \sqrt{48361} from -1.

$$n=\frac{-\sqrt{48361}-1}{6}$$

The equation is now solved.

$$n=\frac{\sqrt{48361}-1}{6} n=\frac{-\sqrt{48361}-1}{6}$$

Steps for Completing the Square

Multiply both sides of the equation by 2.

$$4n+n\left(n-1\right)\times 3=4030$$

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

$$4n+\left(n^{2}-n\right)\times 3=4030$$

Use the distributive property to multiply n^{2}-n by 3.

$$4n+3n^{2}-3n=4030$$

Combine 4n and -3n to get n.

$$n+3n^{2}=4030$$

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.

$$3n^{2}+n=4030$$

Divide both sides by 3.

$$\frac{3n^{2}+n}{3}=\frac{4030}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$n^{2}+\frac{1}{3}n=\frac{4030}{3}$$

Divide \frac{1}{3}, the coefficient of the x term, by 2 to get \frac{1}{6}. Then add the square of \frac{1}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}+\frac{1}{3}n+\left(\frac{1}{6}\right)^{2}=\frac{4030}{3}+\left(\frac{1}{6}\right)^{2}$$

Square \frac{1}{6} by squaring both the numerator and the denominator of the fraction.

$$n^{2}+\frac{1}{3}n+\frac{1}{36}=\frac{4030}{3}+\frac{1}{36}$$

Add \frac{4030}{3} to \frac{1}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$n^{2}+\frac{1}{3}n+\frac{1}{36}=\frac{48361}{36}$$

Factor n^{2}+\frac{1}{3}n+\frac{1}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(n+\frac{1}{6}\right)^{2}=\frac{48361}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n+\frac{1}{6}\right)^{2}}=\sqrt{\frac{48361}{36}}$$

Simplify.

$$n+\frac{1}{6}=\frac{\sqrt{48361}}{6} n+\frac{1}{6}=-\frac{\sqrt{48361}}{6}$$

Subtract \frac{1}{6} from both sides of the equation.

$$n=\frac{\sqrt{48361}-1}{6} n=\frac{-\sqrt{48361}-1}{6}$$