50=128t-16t^2=t(128-16t)

Solution for 50=128t-16t^2=t(128-16t) equation:

50=128t-16t^2=t(128-16t)

We move all terms to the left:

50-(128t-16t^2)=0

We get rid of parentheses

16t^2-128t+50=0

a = 16; b = -128; c = +50;
Δ = b2-4ac
Δ = -1282-4·16·50
Δ = 13184
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:

$\sqrt{\Delta}=\sqrt{13184}=\sqrt{64*206}=\sqrt{64}*\sqrt{206}=8\sqrt{206}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-128)-8\sqrt{206}}{2*16}=\frac{128-8\sqrt{206}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-128)+8\sqrt{206}}{2*16}=\frac{128+8\sqrt{206}}{32}
$