160x+32x^2-192=0

Solution for 160x+32x^2-192=0 equation:

160x+32x^2-192=0

a = 32; b = 160; c = -192;
Δ = b2-4ac
Δ = 1602-4·32·(-192)
Δ = 50176
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{50176}=224$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(160)-224}{2*32}=\frac{-384}{64}
=-6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(160)+224}{2*32}=\frac{64}{64}
=1
$