33=38t^2

Solution for 33=38t^2 equation:

33=38t^2

We move all terms to the left:

33-(38t^2)=0

a = -38; b = 0; c = +33;
Δ = b2-4ac
Δ = 02-4·(-38)·33
Δ = 5016
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:

$\sqrt{\Delta}=\sqrt{5016}=\sqrt{4*1254}=\sqrt{4}*\sqrt{1254}=2\sqrt{1254}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{1254}}{2*-38}=\frac{0-2\sqrt{1254}}{-76}
=-\frac{2\sqrt{1254}}{-76}
=-\frac{\sqrt{1254}}{-38}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{1254}}{2*-38}=\frac{0+2\sqrt{1254}}{-76}
=\frac{2\sqrt{1254}}{-76}
=\frac{\sqrt{1254}}{-38}
$