-16t^2+64t+128t=-3

Solution for -16t^2+64t+128t=-3 equation:

-16t^2+64t+128t=-3

We move all terms to the left:

-16t^2+64t+128t-(-3)=0

We add all the numbers together, and all the variables

-16t^2+192t+3=0

a = -16; b = 192; c = +3;
Δ = b2-4ac
Δ = 1922-4·(-16)·3
Δ = 37056
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:

$\sqrt{\Delta}=\sqrt{37056}=\sqrt{64*579}=\sqrt{64}*\sqrt{579}=8\sqrt{579}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(192)-8\sqrt{579}}{2*-16}=\frac{-192-8\sqrt{579}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(192)+8\sqrt{579}}{2*-16}=\frac{-192+8\sqrt{579}}{-32}
$