v^2+14v+17=0

Solution for v^2+14v+17=0 equation:

v^2+14v+17=0

a = 1; b = 14; c = +17;
Δ = b2-4ac
Δ = 142-4·1·17
Δ = 128
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-8\sqrt{2}}{2*1}=\frac{-14-8\sqrt{2}}{2}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+8\sqrt{2}}{2*1}=\frac{-14+8\sqrt{2}}{2}
$