3n^2+7n-1248=0

Solution for 3n^2+7n-1248=0 equation:

3n^2+7n-1248=0

a = 3; b = 7; c = -1248;
Δ = b2-4ac
Δ = 72-4·3·(-1248)
Δ = 15025
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:

$\sqrt{\Delta}=\sqrt{15025}=\sqrt{25*601}=\sqrt{25}*\sqrt{601}=5\sqrt{601}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-5\sqrt{601}}{2*3}=\frac{-7-5\sqrt{601}}{6}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+5\sqrt{601}}{2*3}=\frac{-7+5\sqrt{601}}{6}
$