2n2+30n-4608=0

Solution for 2n2+30n-4608=0 equation:

2n^2+30n-4608=0

a = 2; b = 30; c = -4608;
Δ = b2-4ac
Δ = 302-4·2·(-4608)
Δ = 37764
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:

$\sqrt{\Delta}=\sqrt{37764}=\sqrt{36*1049}=\sqrt{36}*\sqrt{1049}=6\sqrt{1049}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(30)-6\sqrt{1049}}{2*2}=\frac{-30-6\sqrt{1049}}{4}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(30)+6\sqrt{1049}}{2*2}=\frac{-30+6\sqrt{1049}}{4}
$