# How do you solve for n in (9n+4)+(3n^2 - 0.75) = 180?

Jul 7, 2018

${n}_{1} = \frac{\sqrt{2202}}{6} - \frac{9}{6}$ or ${n}_{2} = - \frac{9}{6} - \frac{\sqrt{2202}}{6}$

#### Explanation:

$3 {n}^{2} + 9 n + 4 - \frac{3}{4} - 180 = 0$

we have

$4 - \frac{3}{4} - 180 = \frac{16 - 3 - 360}{4} = - \frac{707}{4}$

so we have to solve

$3 {n}^{2} + 9 n - \frac{707}{4} = 0$
dividing by $3$ we get

${n}^{2} + 3 n - \frac{707}{12} = 0$

by the quadratic formula we get

${n}_{1 , 2} = - \frac{3}{2} \setminus \pm \sqrt{\frac{367}{6}}$

this is

${n}_{1 , 2} = - \frac{3}{2} \pm \frac{\sqrt{2202}}{6}$

so we get

${n}_{1} = \frac{\sqrt{2202} - 9}{6}$

${n}_{2} = - \frac{1}{6} \left(9 + \sqrt{2202}\right)$