-8t^2+40t+5=0

Solution for -8t^2+40t+5=0 equation:

-8t^2+40t+5=0

a = -8; b = 40; c = +5;
Δ = b2-4ac
Δ = 402-4·(-8)·5
Δ = 1760
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:

$\sqrt{\Delta}=\sqrt{1760}=\sqrt{16*110}=\sqrt{16}*\sqrt{110}=4\sqrt{110}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-4\sqrt{110}}{2*-8}=\frac{-40-4\sqrt{110}}{-16}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+4\sqrt{110}}{2*-8}=\frac{-40+4\sqrt{110}}{-16}
$