5+40t-16t^2=0

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

5+40t-16t^2=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}
$