3n^2+43n-1598=0

Solution for 3n^2+43n-1598=0 equation:

3n^2+43n-1598=0

a = 3; b = 43; c = -1598;
Δ = b2-4ac
Δ = 432-4·3·(-1598)
Δ = 21025
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{21025}=145$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(43)-145}{2*3}=\frac{-188}{6}
=-31+1/3
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(43)+145}{2*3}=\frac{102}{6}
=17
$