2n^2+6n-340=0

Solution for 2n^2+6n-340=0 equation:

2n^2+6n-340=0

a = 2; b = 6; c = -340;
Δ = b2-4ac
Δ = 62-4·2·(-340)
Δ = 2756
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:

$\sqrt{\Delta}=\sqrt{2756}=\sqrt{4*689}=\sqrt{4}*\sqrt{689}=2\sqrt{689}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{689}}{2*2}=\frac{-6-2\sqrt{689}}{4}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{689}}{2*2}=\frac{-6+2\sqrt{689}}{4}
$