(b+-4)(3b+-1)=0

Solution for (b+-4)(3b+-1)=0 equation:

(b+-4)(3b+-1)=0

We add all the numbers together, and all the variables

(b-4)(3b-1)=0

We multiply parentheses ..

(+3b^2-1b-12b+4)=0

We get rid of parentheses

3b^2-1b-12b+4=0

We add all the numbers together, and all the variables

3b^2-13b+4=0

a = 3; b = -13; c = +4;
Δ = b2-4ac
Δ = -132-4·3·4
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-11}{2*3}=\frac{2}{6}
=1/3
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+11}{2*3}=\frac{24}{6}
=4
$