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

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

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

We multiply parentheses ..

(+6b^2-15b+4b-10)=0

We get rid of parentheses

6b^2-15b+4b-10=0

We add all the numbers together, and all the variables

6b^2-11b-10=0

a = 6; b = -11; c = -10;
Δ = b2-4ac
Δ = -112-4·6·(-10)
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{-(-11)-19}{2*6}=\frac{-8}{12}
=-2/3
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+19}{2*6}=\frac{30}{12}
=2+1/2
$