32x^2-48x-32=0

Solution for 32x^2-48x-32=0 equation:

32x^2-48x-32=0

a = 32; b = -48; c = -32;
Δ = b2-4ac
Δ = -482-4·32·(-32)
Δ = 6400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6400}=80$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-80}{2*32}=\frac{-32}{64}
=-1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+80}{2*32}=\frac{128}{64}
=2
$