-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+20t)=0

We get rid of parentheses

16t^2-20t-3=0

a = 16; b = -20; c = -3;
Δ = b2-4ac
Δ = -202-4·16·(-3)
Δ = 592
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:

$\sqrt{\Delta}=\sqrt{592}=\sqrt{16*37}=\sqrt{16}*\sqrt{37}=4\sqrt{37}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{37}}{2*16}=\frac{20-4\sqrt{37}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{37}}{2*16}=\frac{20+4\sqrt{37}}{32}
$