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

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Answer 1 · 2018-07-07T12:10:44

$n_1=sqrt(2202)/6-9/6$ or $n_2=-9/6-sqrt(2202)/6$

Simplifying your equation

$3n^2+9n+4-3/4-180=0$

we have

$4-3/4-180=(16-3-360)/4=-707/4$

so we have to solve

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

$n^2+3n-707/12=0$

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

this is

$n_(1,2)=-3/2pm sqrt(2202)/6$

so we get

$n_1=(sqrt(2202)-9)/6$

$n_2=-1/6(9+sqrt(2202))$

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