33=48t^2

Solution for 33=48t^2 equation:

33=48t^2

We move all terms to the left:

33-(48t^2)=0

a = -48; b = 0; c = +33;
Δ = b2-4ac
Δ = 02-4·(-48)·33
Δ = 6336
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:

$\sqrt{\Delta}=\sqrt{6336}=\sqrt{576*11}=\sqrt{576}*\sqrt{11}=24\sqrt{11}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-24\sqrt{11}}{2*-48}=\frac{0-24\sqrt{11}}{-96}
=-\frac{24\sqrt{11}}{-96}
=-\frac{\sqrt{11}}{-4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+24\sqrt{11}}{2*-48}=\frac{0+24\sqrt{11}}{-96}
=\frac{24\sqrt{11}}{-96}
=\frac{\sqrt{11}}{-4}
$