65t-16t^2=0

Solution for 65t-16t^2=0 equation:

65t-16t^2=0

a = -16; b = 65; c = 0;
Δ = b2-4ac
Δ = 652-4·(-16)·0
Δ = 4225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4225}=65$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(65)-65}{2*-16}=\frac{-130}{-32}
=4+1/16
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(65)+65}{2*-16}=\frac{0}{-32}
=0
$