v^2-16v-48=-4

Solution for v^2-16v-48=-4 equation:

v^2-16v-48=-4

We move all terms to the left:

v^2-16v-48-(-4)=0

We add all the numbers together, and all the variables

v^2-16v-44=0

a = 1; b = -16; c = -44;
Δ = b2-4ac
Δ = -162-4·1·(-44)
Δ = 432
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:

$\sqrt{\Delta}=\sqrt{432}=\sqrt{144*3}=\sqrt{144}*\sqrt{3}=12\sqrt{3}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-12\sqrt{3}}{2*1}=\frac{16-12\sqrt{3}}{2}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+12\sqrt{3}}{2*1}=\frac{16+12\sqrt{3}}{2}
$