3n^2+7n-2050=0

Solution for 3n^2+7n-2050=0 equation:

3n^2+7n-2050=0

a = 3; b = 7; c = -2050;
Δ = b2-4ac
Δ = 72-4·3·(-2050)
Δ = 24649
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{24649}=157$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-157}{2*3}=\frac{-164}{6}
=-27+1/3
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+157}{2*3}=\frac{150}{6}
=25
$