2n^2+4n-109=0

Solution for 2n^2+4n-109=0 equation:

2n^2+4n-109=0

a = 2; b = 4; c = -109;
Δ = b2-4ac
Δ = 42-4·2·(-109)
Δ = 888
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:

$\sqrt{\Delta}=\sqrt{888}=\sqrt{4*222}=\sqrt{4}*\sqrt{222}=2\sqrt{222}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{222}}{2*2}=\frac{-4-2\sqrt{222}}{4}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{222}}{2*2}=\frac{-4+2\sqrt{222}}{4}
$