4q^2-48q+20=0

Solution for 4q^2-48q+20=0 equation:

4q^2-48q+20=0

a = 4; b = -48; c = +20;
Δ = b2-4ac
Δ = -482-4·4·20
Δ = 1984
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1984}=\sqrt{64*31}=\sqrt{64}*\sqrt{31}=8\sqrt{31}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-8\sqrt{31}}{2*4}=\frac{48-8\sqrt{31}}{8}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+8\sqrt{31}}{2*4}=\frac{48+8\sqrt{31}}{8}
$