v^2+8v-36=0

Solution for v^2+8v-36=0 equation:

v^2+8v-36=0

a = 1; b = 8; c = -36;
Δ = b2-4ac
Δ = 82-4·1·(-36)
Δ = 208
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:

$\sqrt{\Delta}=\sqrt{208}=\sqrt{16*13}=\sqrt{16}*\sqrt{13}=4\sqrt{13}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{13}}{2*1}=\frac{-8-4\sqrt{13}}{2}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{13}}{2*1}=\frac{-8+4\sqrt{13}}{2}
$