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