160t-16t^2+2=0

Solution for 160t-16t^2+2=0 equation:

160t-16t^2+2=0

a = -16; b = 160; c = +2;
Δ = b2-4ac
Δ = 1602-4·(-16)·2
Δ = 25728
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{25728}=\sqrt{64*402}=\sqrt{64}*\sqrt{402}=8\sqrt{402}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(160)-8\sqrt{402}}{2*-16}=\frac{-160-8\sqrt{402}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(160)+8\sqrt{402}}{2*-16}=\frac{-160+8\sqrt{402}}{-32}
$