1/4z+6=2/3z+28

Solution for 1/4z+6=2/3z+28 equation:

1/4z+6=2/3z+28

We move all terms to the left:

1/4z+6-(2/3z+28)=0


Domain of the equation: 4z!=0

z!=0/4

z!=0

z∈R



Domain of the equation: 3z+28)!=0

z∈R


We get rid of parentheses

1/4z-2/3z-28+6=0

We calculate fractions


3z/12z^2+(-8z)/12z^2-28+6=0

We add all the numbers together, and all the variables

3z/12z^2+(-8z)/12z^2-22=0

We multiply all the terms by the denominator


3z+(-8z)-22*12z^2=0

Wy multiply elements

-264z^2+3z+(-8z)=0

We get rid of parentheses

-264z^2+3z-8z=0

We add all the numbers together, and all the variables

-264z^2-5z=0

a = -264; b = -5; c = 0;
Δ = b2-4ac
Δ = -52-4·(-264)·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-5}{2*-264}=\frac{0}{-528}
=0
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5}{2*-264}=\frac{10}{-528}
=-5/264
$