(3b-1)(2b+5)=3

Solution for (3b-1)(2b+5)=3 equation:

(3b-1)(2b+5)=3

We move all terms to the left:

(3b-1)(2b+5)-(3)=0

We multiply parentheses ..

(+6b^2+15b-2b-5)-3=0

We get rid of parentheses

6b^2+15b-2b-5-3=0

We add all the numbers together, and all the variables

6b^2+13b-8=0

a = 6; b = 13; c = -8;
Δ = b2-4ac
Δ = 132-4·6·(-8)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-19}{2*6}=\frac{-32}{12}
=-2+2/3
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+19}{2*6}=\frac{6}{12}
=1/2
$