(5b+4)(3b+1)=0

Solution for (5b+4)(3b+1)=0 equation:

(5b+4)(3b+1)=0

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

15b^2+17b+4=0

a = 15; b = 17; c = +4;
Δ = b2-4ac
Δ = 172-4·15·4
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-7}{2*15}=\frac{-24}{30}
=-4/5
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+7}{2*15}=\frac{-10}{30}
=-1/3
$