4n2+2n=132

Solution for 4n2+2n=132 equation:

4n^2+2n=132

We move all terms to the left:

4n^2+2n-(132)=0

a = 4; b = 2; c = -132;
Δ = b2-4ac
Δ = 22-4·4·(-132)
Δ = 2116
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2116}=46$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-46}{2*4}=\frac{-48}{8}
=-6
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+46}{2*4}=\frac{44}{8}
=5+1/2
$