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

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

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

We multiply parentheses ..

(+12b^2+16b+3b+4)=0

We get rid of parentheses

12b^2+16b+3b+4=0

We add all the numbers together, and all the variables

12b^2+19b+4=0

a = 12; b = 19; c = +4;
Δ = b2-4ac
Δ = 192-4·12·4
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-13}{2*12}=\frac{-32}{24}
=-1+1/3
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+13}{2*12}=\frac{-6}{24}
=-1/4
$