32b^2-48b^2-2b+3=0

Solution for 32b^2-48b^2-2b+3=0 equation:

32b^2-48b^2-2b+3=0

We add all the numbers together, and all the variables

-16b^2-2b+3=0

a = -16; b = -2; c = +3;
Δ = b2-4ac
Δ = -22-4·(-16)·3
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-14}{2*-16}=\frac{-12}{-32}
=3/8
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+14}{2*-16}=\frac{16}{-32}
=-1/2
$