v^2-14v-3=0

Solution for v^2-14v-3=0 equation:

v^2-14v-3=0

a = 1; b = -14; c = -3;
Δ = b2-4ac
Δ = -142-4·1·(-3)
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{-(-14)-4\sqrt{13}}{2*1}=\frac{14-4\sqrt{13}}{2}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+4\sqrt{13}}{2*1}=\frac{14+4\sqrt{13}}{2}
$