z/3z+8=4z+9

Solution for z/3z+8=4z+9 equation:

z/3z+8=4z+9

We move all terms to the left:

z/3z+8-(4z+9)=0


Domain of the equation: 3z!=0

z!=0/3

z!=0

z∈R


We get rid of parentheses

z/3z-4z-9+8=0

We multiply all the terms by the denominator


z-4z*3z-9*3z+8*3z=0

Wy multiply elements

-12z^2+z-27z+24z=0

We add all the numbers together, and all the variables

-12z^2-2z=0

a = -12; b = -2; c = 0;
Δ = b2-4ac
Δ = -22-4·(-12)·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2}{2*-12}=\frac{0}{-24}
=0
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2}{2*-12}=\frac{4}{-24}
=-1/6
$