-16q^2+128q=192

Solution for -16q^2+128q=192 equation:

-16q^2+128q=192

We move all terms to the left:

-16q^2+128q-(192)=0

a = -16; b = 128; c = -192;
Δ = b2-4ac
Δ = 1282-4·(-16)·(-192)
Δ = 4096
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4096}=64$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(128)-64}{2*-16}=\frac{-192}{-32}
=+6
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(128)+64}{2*-16}=\frac{-64}{-32}
=+2
$