2n^2+4n=286

Solution for 2n^2+4n=286 equation:

2n^2+4n=286

We move all terms to the left:

2n^2+4n-(286)=0

a = 2; b = 4; c = -286;
Δ = b2-4ac
Δ = 42-4·2·(-286)
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2304}=48$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-48}{2*2}=\frac{-52}{4}
=-13
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+48}{2*2}=\frac{44}{4}
=11
$