2t2-16t+40-80=0

Solution for 2t2-16t+40-80=0 equation:

2t^2-16t+40-80=0

We add all the numbers together, and all the variables

2t^2-16t-40=0

a = 2; b = -16; c = -40;
Δ = b2-4ac
Δ = -162-4·2·(-40)
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-24}{2*2}=\frac{-8}{4}
=-2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+24}{2*2}=\frac{40}{4}
=10
$