n+2=3n^2+24n+30

Solution for n+2=3n^2+24n+30 equation:

n+2=3n^2+24n+30

We move all terms to the left:

n+2-(3n^2+24n+30)=0

We get rid of parentheses

-3n^2+n-24n-30+2=0

We add all the numbers together, and all the variables

-3n^2-23n-28=0

a = -3; b = -23; c = -28;
Δ = b2-4ac
Δ = -232-4·(-3)·(-28)
Δ = 193
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-\sqrt{193}}{2*-3}=\frac{23-\sqrt{193}}{-6}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+\sqrt{193}}{2*-3}=\frac{23+\sqrt{193}}{-6}
$