4z(3z+1)=9z+3

Solution for 4z(3z+1)=9z+3 equation:

4z(3z+1)=9z+3

We move all terms to the left:

4z(3z+1)-(9z+3)=0

We multiply parentheses

12z^2+4z-(9z+3)=0

We get rid of parentheses

12z^2+4z-9z-3=0

We add all the numbers together, and all the variables

12z^2-5z-3=0

a = 12; b = -5; c = -3;
Δ = b2-4ac
Δ = -52-4·12·(-3)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-13}{2*12}=\frac{-8}{24}
=-1/3
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+13}{2*12}=\frac{18}{24}
=3/4
$