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

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

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

6b^2+17b+5=0

a = 6; b = 17; c = +5;
Δ = b2-4ac
Δ = 172-4·6·5
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-13}{2*6}=\frac{-30}{12}
=-2+1/2
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+13}{2*6}=\frac{-4}{12}
=-1/3
$