t^2+40t-80=0

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

t^2+40t-80=0

a = 1; b = 40; c = -80;
Δ = b2-4ac
Δ = 402-4·1·(-80)
Δ = 1920
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:

$\sqrt{\Delta}=\sqrt{1920}=\sqrt{64*30}=\sqrt{64}*\sqrt{30}=8\sqrt{30}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-8\sqrt{30}}{2*1}=\frac{-40-8\sqrt{30}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+8\sqrt{30}}{2*1}=\frac{-40+8\sqrt{30}}{2}
$