20=-16t2+30t+8

Solution for 20=-16t2+30t+8 equation:

20=-16t^2+30t+8

We move all terms to the left:

20-(-16t^2+30t+8)=0

We get rid of parentheses

16t^2-30t-8+20=0

We add all the numbers together, and all the variables

16t^2-30t+12=0

a = 16; b = -30; c = +12;
Δ = b2-4ac
Δ = -302-4·16·12
Δ = 132
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:

$\sqrt{\Delta}=\sqrt{132}=\sqrt{4*33}=\sqrt{4}*\sqrt{33}=2\sqrt{33}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-30)-2\sqrt{33}}{2*16}=\frac{30-2\sqrt{33}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-30)+2\sqrt{33}}{2*16}=\frac{30+2\sqrt{33}}{32}
$