6n^2+4n=1020

Solution for 6n^2+4n=1020 equation:

6n^2+4n=1020

We move all terms to the left:

6n^2+4n-(1020)=0

a = 6; b = 4; c = -1020;
Δ = b2-4ac
Δ = 42-4·6·(-1020)
Δ = 24496
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:

$\sqrt{\Delta}=\sqrt{24496}=\sqrt{16*1531}=\sqrt{16}*\sqrt{1531}=4\sqrt{1531}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{1531}}{2*6}=\frac{-4-4\sqrt{1531}}{12}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{1531}}{2*6}=\frac{-4+4\sqrt{1531}}{12}
$