2n^2+8n-93=0

Solution for 2n^2+8n-93=0 equation:

2n^2+8n-93=0

a = 2; b = 8; c = -93;
Δ = b2-4ac
Δ = 82-4·2·(-93)
Δ = 808
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:

$\sqrt{\Delta}=\sqrt{808}=\sqrt{4*202}=\sqrt{4}*\sqrt{202}=2\sqrt{202}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{202}}{2*2}=\frac{-8-2\sqrt{202}}{4}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{202}}{2*2}=\frac{-8+2\sqrt{202}}{4}
$