4v^2-32v+48=0

Solution for 4v^2-32v+48=0 equation:

4v^2-32v+48=0

a = 4; b = -32; c = +48;
Δ = b2-4ac
Δ = -322-4·4·48
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{-(-32)-16}{2*4}=\frac{16}{8}
=2
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+16}{2*4}=\frac{48}{8}
=6
$