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

Question

2526 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-07T12:10:44

$n_{1} = \frac{\sqrt{2202}}{6} - \frac{9}{6}$ $n_1=sqrt(2202)/6-9/6$ or $n_{2} = - \frac{9}{6} - \frac{\sqrt{2202}}{6}$ $n_2=-9/6-sqrt(2202)/6$

Simplifying your equation

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

we have

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

so we have to solve

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

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

$n_{1, 2} = - \frac{3}{2} \pm \sqrt{\frac{367}{6}}$ $n_(1,2)=-3/2\pm sqrt(367/6)$

this is

$n_{1, 2} = - \frac{3}{2} \pm \frac{\sqrt{2202}}{6}$ $n_(1,2)=-3/2pm sqrt(2202)/6$

so we get

$n_{1} = \frac{\sqrt{2202} - 9}{6}$ $n_1=(sqrt(2202)-9)/6$

$n_{2} = - \frac{1}{6} (9 + \sqrt{2202})$ $n_2=-1/6(9+sqrt(2202))$

How do you solve for n in (9n+4)+(3n2−0.75)=180(9n+4)+(3n^2 - 0.75) = 180?