(b-2)(3b+1)=0

Solution for (b-2)(3b+1)=0 equation:

(b-2)(3b+1)=0

We multiply parentheses ..

(+3b^2+b-6b-2)=0

We get rid of parentheses

3b^2+b-6b-2=0

We add all the numbers together, and all the variables

3b^2-5b-2=0

a = 3; b = -5; c = -2;
Δ = b2-4ac
Δ = -52-4·3·(-2)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-7}{2*3}=\frac{-2}{6}
=-1/3
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+7}{2*3}=\frac{12}{6}
=2
$