v^2+12v-99=0

Solution for v^2+12v-99=0 equation:

v^2+12v-99=0

a = 1; b = 12; c = -99;
Δ = b2-4ac
Δ = 122-4·1·(-99)
Δ = 540
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:

$\sqrt{\Delta}=\sqrt{540}=\sqrt{36*15}=\sqrt{36}*\sqrt{15}=6\sqrt{15}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-6\sqrt{15}}{2*1}=\frac{-12-6\sqrt{15}}{2}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+6\sqrt{15}}{2*1}=\frac{-12+6\sqrt{15}}{2}
$