15t^2+40t-80=0

Solution for 15t^2+40t-80=0 equation:

15t^2+40t-80=0

a = 15; b = 40; c = -80;
Δ = b2-4ac
Δ = 402-4·15·(-80)
Δ = 6400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6400}=80$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-80}{2*15}=\frac{-120}{30}
=-4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+80}{2*15}=\frac{40}{30}
=1+1/3
$