20=-16t^2+10t+40

Solution for 20=-16t^2+10t+40 equation:

20=-16t^2+10t+40

We move all terms to the left:

20-(-16t^2+10t+40)=0

We get rid of parentheses

16t^2-10t-40+20=0

We add all the numbers together, and all the variables

16t^2-10t-20=0

a = 16; b = -10; c = -20;
Δ = b2-4ac
Δ = -102-4·16·(-20)
Δ = 1380
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:

$\sqrt{\Delta}=\sqrt{1380}=\sqrt{4*345}=\sqrt{4}*\sqrt{345}=2\sqrt{345}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{345}}{2*16}=\frac{10-2\sqrt{345}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{345}}{2*16}=\frac{10+2\sqrt{345}}{32}
$