-3=-16t^2=20t

Solution for -3=-16t^2=20t equation:

-3=-16t^2=20t

We move all terms to the left:

-3-(-16t^2)=0

We get rid of parentheses

16t^2-3=0

a = 16; b = 0; c = -3;
Δ = b2-4ac
Δ = 02-4·16·(-3)
Δ = 192
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:

$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{3}}{2*16}=\frac{0-8\sqrt{3}}{32}
=-\frac{8\sqrt{3}}{32}
=-\frac{\sqrt{3}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{3}}{2*16}=\frac{0+8\sqrt{3}}{32}
=\frac{8\sqrt{3}}{32}
=\frac{\sqrt{3}}{4}
$