v^2=-8v+105

Solution for v^2=-8v+105 equation:

v^2=-8v+105

We move all terms to the left:

v^2-(-8v+105)=0

We get rid of parentheses

v^2+8v-105=0

a = 1; b = 8; c = -105;
Δ = b2-4ac
Δ = 82-4·1·(-105)
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-22}{2*1}=\frac{-30}{2}
=-15
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+22}{2*1}=\frac{14}{2}
=7
$