20=48t^2

Solution for 20=48t^2 equation:

20=48t^2

We move all terms to the left:

20-(48t^2)=0

a = -48; b = 0; c = +20;
Δ = b2-4ac
Δ = 02-4·(-48)·20
Δ = 3840
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{3840}=\sqrt{256*15}=\sqrt{256}*\sqrt{15}=16\sqrt{15}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{15}}{2*-48}=\frac{0-16\sqrt{15}}{-96}
=-\frac{16\sqrt{15}}{-96}
=-\frac{\sqrt{15}}{-6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{15}}{2*-48}=\frac{0+16\sqrt{15}}{-96}
=\frac{16\sqrt{15}}{-96}
=\frac{\sqrt{15}}{-6}
$