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

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

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

We multiply parentheses ..

(+6b^2-9b+10b-15)=0

We get rid of parentheses

6b^2-9b+10b-15=0

We add all the numbers together, and all the variables

6b^2+b-15=0

a = 6; b = 1; c = -15;
Δ = b2-4ac
Δ = 12-4·6·(-15)
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{-(1)-19}{2*6}=\frac{-20}{12}
=-1+2/3
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+19}{2*6}=\frac{18}{12}
=1+1/2
$