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 = 1440
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:

$\sqrt{\Delta}=\sqrt{1440}=\sqrt{144*10}=\sqrt{144}*\sqrt{10}=12\sqrt{10}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-12\sqrt{10}}{2*-8}=\frac{40-12\sqrt{10}}{-16}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+12\sqrt{10}}{2*-8}=\frac{40+12\sqrt{10}}{-16}
$