-16t2+80t=64+0

Solution for -16t2+80t=64+0 equation:

-16t^2+80t=64+0

We move all terms to the left:

-16t^2+80t-(64+0)=0

We add all the numbers together, and all the variables

-16t^2+80t-64=0

a = -16; b = 80; c = -64;
Δ = b2-4ac
Δ = 802-4·(-16)·(-64)
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2304}=48$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(80)-48}{2*-16}=\frac{-128}{-32}
=+4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(80)+48}{2*-16}=\frac{-32}{-32}
=1
$