0=40t^2+80t-100

Solution for 0=40t^2+80t-100 equation:

0=40t^2+80t-100

We move all terms to the left:

0-(40t^2+80t-100)=0

We add all the numbers together, and all the variables

-(40t^2+80t-100)=0

We get rid of parentheses

-40t^2-80t+100=0

a = -40; b = -80; c = +100;
Δ = b2-4ac
Δ = -802-4·(-40)·100
Δ = 22400
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:

$\sqrt{\Delta}=\sqrt{22400}=\sqrt{1600*14}=\sqrt{1600}*\sqrt{14}=40\sqrt{14}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-80)-40\sqrt{14}}{2*-40}=\frac{80-40\sqrt{14}}{-80}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-80)+40\sqrt{14}}{2*-40}=\frac{80+40\sqrt{14}}{-80}
$