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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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}
$