-15t^2+40t+15=0

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

-15t^2+40t+15=0

a = -15; b = 40; c = +15;
Δ = b2-4ac
Δ = 402-4·(-15)·15
Δ = 2500
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2500}=50$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-50}{2*-15}=\frac{-90}{-30}
=+3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+50}{2*-15}=\frac{10}{-30}
=-1/3
$