-16t^2+64t+0=0

Solution for -16t^2+64t+0=0 equation:

-16t^2+64t+0=0

We add all the numbers together, and all the variables

-16t^2+64t=0

a = -16; b = 64; c = 0;
Δ = b2-4ac
Δ = 642-4·(-16)·0
Δ = 4096
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4096}=64$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(64)-64}{2*-16}=\frac{-128}{-32}
=+4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(64)+64}{2*-16}=\frac{0}{-32}
=0
$