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

Solution for 0=(3b-1)(2b+5) equation:

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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


We calculate terms in parentheses: -((+6b^2+15b-2b-5)), so:

(+6b^2+15b-2b-5)

We get rid of parentheses

6b^2+15b-2b-5

We add all the numbers together, and all the variables

6b^2+13b-5

Back to the equation:

-(6b^2+13b-5)


We get rid of parentheses

-6b^2-13b+5=0

a = -6; b = -13; c = +5;
Δ = b2-4ac
Δ = -132-4·(-6)·5
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-17}{2*-6}=\frac{-4}{-12}
=1/3
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+17}{2*-6}=\frac{30}{-12}
=-2+1/2
$