-16t^2+52t=0

Solution for -16t^2+52t=0 equation:

-16t^2+52t=0

a = -16; b = 52; c = 0;
Δ = b2-4ac
Δ = 522-4·(-16)·0
Δ = 2704
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2704}=52$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(52)-52}{2*-16}=\frac{-104}{-32}
=3+1/4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(52)+52}{2*-16}=\frac{0}{-32}
=0
$