4v^2+12v-20=-4

Solution for 4v^2+12v-20=-4 equation:

4v^2+12v-20=-4

We move all terms to the left:

4v^2+12v-20-(-4)=0

We add all the numbers together, and all the variables

4v^2+12v-16=0

a = 4; b = 12; c = -16;
Δ = b2-4ac
Δ = 122-4·4·(-16)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-20}{2*4}=\frac{-32}{8}
=-4
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+20}{2*4}=\frac{8}{8}
=1
$