(5b-4)(b-3)=0

Solution for (5b-4)(b-3)=0 equation:

(5b-4)(b-3)=0

We multiply parentheses ..

(+5b^2-15b-4b+12)=0

We get rid of parentheses

5b^2-15b-4b+12=0

We add all the numbers together, and all the variables

5b^2-19b+12=0

a = 5; b = -19; c = +12;
Δ = b2-4ac
Δ = -192-4·5·12
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-11}{2*5}=\frac{8}{10}
=4/5
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+11}{2*5}=\frac{30}{10}
=3
$