-15t^2+32t=0

Solution for -15t^2+32t=0 equation:

-15t^2+32t=0

a = -15; b = 32; c = 0;
Δ = b2-4ac
Δ = 322-4·(-15)·0
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-32}{2*-15}=\frac{-64}{-30}
=2+2/15
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+32}{2*-15}=\frac{0}{-30}
=0
$