-16t^2+160t=256

Solution for -16t^2+160t=256 equation:

-16t^2+160t=256

We move all terms to the left:

-16t^2+160t-(256)=0

a = -16; b = 160; c = -256;
Δ = b2-4ac
Δ = 1602-4·(-16)·(-256)
Δ = 9216
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9216}=96$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(160)-96}{2*-16}=\frac{-256}{-32}
=+8
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(160)+96}{2*-16}=\frac{-64}{-32}
=+2
$