3n2+35n-852=0

Solution for 3n2+35n-852=0 equation:

3n^2+35n-852=0

a = 3; b = 35; c = -852;
Δ = b2-4ac
Δ = 352-4·3·(-852)
Δ = 11449
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{11449}=107$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(35)-107}{2*3}=\frac{-142}{6}
=-23+2/3
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(35)+107}{2*3}=\frac{72}{6}
=12
$