n^2+2n^2-9n-18=0

Solution for n^2+2n^2-9n-18=0 equation:

n^2+2n^2-9n-18=0

We add all the numbers together, and all the variables

3n^2-9n-18=0

a = 3; b = -9; c = -18;
Δ = b2-4ac
Δ = -92-4·3·(-18)
Δ = 297
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{297}=\sqrt{9*33}=\sqrt{9}*\sqrt{33}=3\sqrt{33}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-3\sqrt{33}}{2*3}=\frac{9-3\sqrt{33}}{6}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+3\sqrt{33}}{2*3}=\frac{9+3\sqrt{33}}{6}
$