v^2-16v-23=0

Solution for v^2-16v-23=0 equation:

v^2-16v-23=0

a = 1; b = -16; c = -23;
Δ = b2-4ac
Δ = -162-4·1·(-23)
Δ = 348
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:

$\sqrt{\Delta}=\sqrt{348}=\sqrt{4*87}=\sqrt{4}*\sqrt{87}=2\sqrt{87}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-2\sqrt{87}}{2*1}=\frac{16-2\sqrt{87}}{2}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+2\sqrt{87}}{2*1}=\frac{16+2\sqrt{87}}{2}
$