0=-16t^2+(40)t+5

Solution for 0=-16t^2+(40)t+5 equation:

0=-16t^2+(40)t+5

We move all terms to the left:

0-(-16t^2+(40)t+5)=0

We add all the numbers together, and all the variables

-(-16t^2+40t+5)=0

We get rid of parentheses

16t^2-40t-5=0

a = 16; b = -40; c = -5;
Δ = b2-4ac
Δ = -402-4·16·(-5)
Δ = 1920
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:

$\sqrt{\Delta}=\sqrt{1920}=\sqrt{64*30}=\sqrt{64}*\sqrt{30}=8\sqrt{30}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-8\sqrt{30}}{2*16}=\frac{40-8\sqrt{30}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+8\sqrt{30}}{2*16}=\frac{40+8\sqrt{30}}{32}
$