v^2+12v+34=0

Solution for v^2+12v+34=0 equation:

v^2+12v+34=0

a = 1; b = 12; c = +34;
Δ = b2-4ac
Δ = 122-4·1·34
Δ = 8
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:

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