40t^2+80t-100=0

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

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}
$