8=-16t^2+20t+4

Solution for 8=-16t^2+20t+4 equation:

8=-16t^2+20t+4

We move all terms to the left:

8-(-16t^2+20t+4)=0

We get rid of parentheses

16t^2-20t-4+8=0

We add all the numbers together, and all the variables

16t^2-20t+4=0

a = 16; b = -20; c = +4;
Δ = b2-4ac
Δ = -202-4·16·4
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-12}{2*16}=\frac{8}{32}
=1/4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+12}{2*16}=\frac{32}{32}
=1
$