v^2-10v-48=-8

Solution for v^2-10v-48=-8 equation:

v^2-10v-48=-8

We move all terms to the left:

v^2-10v-48-(-8)=0

We add all the numbers together, and all the variables

v^2-10v-40=0

a = 1; b = -10; c = -40;
Δ = b2-4ac
Δ = -102-4·1·(-40)
Δ = 260
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:

$\sqrt{\Delta}=\sqrt{260}=\sqrt{4*65}=\sqrt{4}*\sqrt{65}=2\sqrt{65}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{65}}{2*1}=\frac{10-2\sqrt{65}}{2}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{65}}{2*1}=\frac{10+2\sqrt{65}}{2}
$