3n^2+2n-200=0

Solution for 3n^2+2n-200=0 equation:

3n^2+2n-200=0

a = 3; b = 2; c = -200;
Δ = b2-4ac
Δ = 22-4·3·(-200)
Δ = 2404
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:

$\sqrt{\Delta}=\sqrt{2404}=\sqrt{4*601}=\sqrt{4}*\sqrt{601}=2\sqrt{601}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{601}}{2*3}=\frac{-2-2\sqrt{601}}{6}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{601}}{2*3}=\frac{-2+2\sqrt{601}}{6}
$