3n^2+24n-18=-9

Solution for 3n^2+24n-18=-9 equation:

3n^2+24n-18=-9

We move all terms to the left:

3n^2+24n-18-(-9)=0

We add all the numbers together, and all the variables

3n^2+24n-9=0

a = 3; b = 24; c = -9;
Δ = b2-4ac
Δ = 242-4·3·(-9)
Δ = 684
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:

$\sqrt{\Delta}=\sqrt{684}=\sqrt{36*19}=\sqrt{36}*\sqrt{19}=6\sqrt{19}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-6\sqrt{19}}{2*3}=\frac{-24-6\sqrt{19}}{6}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+6\sqrt{19}}{2*3}=\frac{-24+6\sqrt{19}}{6}
$