v^2+8v-62=0

Solution for v^2+8v-62=0 equation:

v^2+8v-62=0

a = 1; b = 8; c = -62;
Δ = b2-4ac
Δ = 82-4·1·(-62)
Δ = 312
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:

$\sqrt{\Delta}=\sqrt{312}=\sqrt{4*78}=\sqrt{4}*\sqrt{78}=2\sqrt{78}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{78}}{2*1}=\frac{-8-2\sqrt{78}}{2}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{78}}{2*1}=\frac{-8+2\sqrt{78}}{2}
$