2n^2+4n=28

Solution for 2n^2+4n=28 equation:

2n^2+4n=28

We move all terms to the left:

2n^2+4n-(28)=0

a = 2; b = 4; c = -28;
Δ = b2-4ac
Δ = 42-4·2·(-28)
Δ = 240
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:

$\sqrt{\Delta}=\sqrt{240}=\sqrt{16*15}=\sqrt{16}*\sqrt{15}=4\sqrt{15}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{15}}{2*2}=\frac{-4-4\sqrt{15}}{4}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{15}}{2*2}=\frac{-4+4\sqrt{15}}{4}
$