-16t^2+128t=-50

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

-16t^2+128t=-50

We move all terms to the left:

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

We add all the numbers together, and all the variables

-16t^2+128t+50=0

a = -16; b = 128; c = +50;
Δ = b2-4ac
Δ = 1282-4·(-16)·50
Δ = 19584
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:

$\sqrt{\Delta}=\sqrt{19584}=\sqrt{576*34}=\sqrt{576}*\sqrt{34}=24\sqrt{34}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(128)-24\sqrt{34}}{2*-16}=\frac{-128-24\sqrt{34}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(128)+24\sqrt{34}}{2*-16}=\frac{-128+24\sqrt{34}}{-32}
$