2n^2+12n+17=0

Solution for 2n^2+12n+17=0 equation:

2n^2+12n+17=0

a = 2; b = 12; c = +17;
Δ = b2-4ac
Δ = 122-4·2·17
Δ = 8
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:

$\sqrt{\Delta}=\sqrt{8}=\sqrt{4*2}=\sqrt{4}*\sqrt{2}=2\sqrt{2}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{2}}{2*2}=\frac{-12-2\sqrt{2}}{4}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{2}}{2*2}=\frac{-12+2\sqrt{2}}{4}
$