5=16t^2+20t

Solution for 5=16t^2+20t equation:

5=16t^2+20t

We move all terms to the left:

5-(16t^2+20t)=0

We get rid of parentheses

-16t^2-20t+5=0

a = -16; b = -20; c = +5;
Δ = b2-4ac
Δ = -202-4·(-16)·5
Δ = 720
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:

$\sqrt{\Delta}=\sqrt{720}=\sqrt{144*5}=\sqrt{144}*\sqrt{5}=12\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-12\sqrt{5}}{2*-16}=\frac{20-12\sqrt{5}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+12\sqrt{5}}{2*-16}=\frac{20+12\sqrt{5}}{-32}
$