v^2+14v-94=0

Solution for v^2+14v-94=0 equation:

v^2+14v-94=0

a = 1; b = 14; c = -94;
Δ = b2-4ac
Δ = 142-4·1·(-94)
Δ = 572
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:

$\sqrt{\Delta}=\sqrt{572}=\sqrt{4*143}=\sqrt{4}*\sqrt{143}=2\sqrt{143}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{143}}{2*1}=\frac{-14-2\sqrt{143}}{2}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{143}}{2*1}=\frac{-14+2\sqrt{143}}{2}
$