v^2-10v-39=0

Solution for v^2-10v-39=0 equation:

v^2-10v-39=0

a = 1; b = -10; c = -39;
Δ = b2-4ac
Δ = -102-4·1·(-39)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-16}{2*1}=\frac{-6}{2}
=-3
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+16}{2*1}=\frac{26}{2}
=13
$