Solve 630=n(0+5n+2)/2

$$630 = \frac { n ( 0 + 5 n + 2 ) } { 2 }$$

Solve for n

$$n = \frac{\sqrt{6301} - 1}{5} \approx 15.675767698$$

$$n=\frac{-\sqrt{6301}-1}{5}\approx -16.075767698$$

Steps Using the Quadratic Formula

Multiply both sides by 2.

$$630\times 2=n\left(0+5n+2\right)$$

Multiply 630 and 2 to get 1260.

$$1260=n\left(0+5n+2\right)$$

Add 0 and 2 to get 2.

$$1260=n\left(2+5n\right)$$

Use the distributive property to multiply n by 2+5n.

$$1260=2n+5n^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$2n+5n^{2}=1260$$

Subtract 1260 from both sides.

$$2n+5n^{2}-1260=0$$

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.

$$5n^{2}+2n-1260=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, 2 for b, and -1260 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-2±\sqrt{2^{2}-4\times 5\left(-1260\right)}}{2\times 5}$$

Square 2.

$$n=\frac{-2±\sqrt{4-4\times 5\left(-1260\right)}}{2\times 5}$$

Multiply -4 times 5.

$$n=\frac{-2±\sqrt{4-20\left(-1260\right)}}{2\times 5}$$

Multiply -20 times -1260.

$$n=\frac{-2±\sqrt{4+25200}}{2\times 5}$$

Add 4 to 25200.

$$n=\frac{-2±\sqrt{25204}}{2\times 5}$$

Take the square root of 25204.

$$n=\frac{-2±2\sqrt{6301}}{2\times 5}$$

Multiply 2 times 5.

$$n=\frac{-2±2\sqrt{6301}}{10}$$

Now solve the equation n=\frac{-2±2\sqrt{6301}}{10} when ± is plus. Add -2 to 2\sqrt{6301}.

$$n=\frac{2\sqrt{6301}-2}{10}$$

Divide -2+2\sqrt{6301} by 10.

$$n=\frac{\sqrt{6301}-1}{5}$$

Now solve the equation n=\frac{-2±2\sqrt{6301}}{10} when ± is minus. Subtract 2\sqrt{6301} from -2.

$$n=\frac{-2\sqrt{6301}-2}{10}$$

Divide -2-2\sqrt{6301} by 10.

$$n=\frac{-\sqrt{6301}-1}{5}$$

The equation is now solved.

$$n=\frac{\sqrt{6301}-1}{5} n=\frac{-\sqrt{6301}-1}{5}$$

Steps for Completing the Square

Multiply both sides by 2.

$$630\times 2=n\left(0+5n+2\right)$$

Multiply 630 and 2 to get 1260.

$$1260=n\left(0+5n+2\right)$$

Add 0 and 2 to get 2.

$$1260=n\left(2+5n\right)$$

Use the distributive property to multiply n by 2+5n.

$$1260=2n+5n^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$2n+5n^{2}=1260$$

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.

$$5n^{2}+2n=1260$$

Divide both sides by 5.

$$\frac{5n^{2}+2n}{5}=\frac{1260}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$n^{2}+\frac{2}{5}n=\frac{1260}{5}$$

Divide 1260 by 5.

$$n^{2}+\frac{2}{5}n=252$$

Divide \frac{2}{5}, the coefficient of the x term, by 2 to get \frac{1}{5}. Then add the square of \frac{1}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}+\frac{2}{5}n+\left(\frac{1}{5}\right)^{2}=252+\left(\frac{1}{5}\right)^{2}$$

Square \frac{1}{5} by squaring both the numerator and the denominator of the fraction.

$$n^{2}+\frac{2}{5}n+\frac{1}{25}=252+\frac{1}{25}$$

Add 252 to \frac{1}{25}.

$$n^{2}+\frac{2}{5}n+\frac{1}{25}=\frac{6301}{25}$$

Factor n^{2}+\frac{2}{5}n+\frac{1}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(n+\frac{1}{5}\right)^{2}=\frac{6301}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n+\frac{1}{5}\right)^{2}}=\sqrt{\frac{6301}{25}}$$

Simplify.

$$n+\frac{1}{5}=\frac{\sqrt{6301}}{5} n+\frac{1}{5}=-\frac{\sqrt{6301}}{5}$$

Subtract \frac{1}{5} from both sides of the equation.

$$n=\frac{\sqrt{6301}-1}{5} n=\frac{-\sqrt{6301}-1}{5}$$