0=-16t^2+80t+40

Solution for 0=-16t^2+80t+40 equation:

0=-16t^2+80t+40

We move all terms to the left:

0-(-16t^2+80t+40)=0

We add all the numbers together, and all the variables

-(-16t^2+80t+40)=0

We get rid of parentheses

16t^2-80t-40=0

a = 16; b = -80; c = -40;
Δ = b2-4ac
Δ = -802-4·16·(-40)
Δ = 8960
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:

$\sqrt{\Delta}=\sqrt{8960}=\sqrt{256*35}=\sqrt{256}*\sqrt{35}=16\sqrt{35}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-80)-16\sqrt{35}}{2*16}=\frac{80-16\sqrt{35}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-80)+16\sqrt{35}}{2*16}=\frac{80+16\sqrt{35}}{32}
$