2n^2+4n=26

Solution for 2n^2+4n=26 equation:

2n^2+4n=26

We move all terms to the left:

2n^2+4n-(26)=0

a = 2; b = 4; c = -26;
Δ = b2-4ac
Δ = 42-4·2·(-26)
Δ = 224
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:

$\sqrt{\Delta}=\sqrt{224}=\sqrt{16*14}=\sqrt{16}*\sqrt{14}=4\sqrt{14}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{14}}{2*2}=\frac{-4-4\sqrt{14}}{4}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{14}}{2*2}=\frac{-4+4\sqrt{14}}{4}
$