32n2+n-4608=0

Solution for 32n2+n-4608=0 equation:

32n^2+n-4608=0

a = 32; b = 1; c = -4608;
Δ = b2-4ac
Δ = 12-4·32·(-4608)
Δ = 589825
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:

$\sqrt{\Delta}=\sqrt{589825}=\sqrt{25*23593}=\sqrt{25}*\sqrt{23593}=5\sqrt{23593}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-5\sqrt{23593}}{2*32}=\frac{-1-5\sqrt{23593}}{64}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+5\sqrt{23593}}{2*32}=\frac{-1+5\sqrt{23593}}{64}
$