-15t^2+20t=0

Solution for -15t^2+20t=0 equation:

-15t^2+20t=0

a = -15; b = 20; c = 0;
Δ = b2-4ac
Δ = 202-4·(-15)·0
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-20}{2*-15}=\frac{-40}{-30}
=1+1/3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+20}{2*-15}=\frac{0}{-30}
=0
$