1/3z-3=2-7/4z

Solution for 1/3z-3=2-7/4z equation:

1/3z-3=2-7/4z

We move all terms to the left:

1/3z-3-(2-7/4z)=0


Domain of the equation: 3z!=0

z!=0/3

z!=0

z∈R



Domain of the equation: 4z)!=0

z!=0/1

z!=0

z∈R


We add all the numbers together, and all the variables

1/3z-(-7/4z+2)-3=0

We get rid of parentheses

1/3z+7/4z-2-3=0

We calculate fractions


4z/12z^2+21z/12z^2-2-3=0

We add all the numbers together, and all the variables

4z/12z^2+21z/12z^2-5=0

We multiply all the terms by the denominator


4z+21z-5*12z^2=0

We add all the numbers together, and all the variables

25z-5*12z^2=0

Wy multiply elements

-60z^2+25z=0

a = -60; b = 25; c = 0;
Δ = b2-4ac
Δ = 252-4·(-60)·0
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-25}{2*-60}=\frac{-50}{-120}
=5/12
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+25}{2*-60}=\frac{0}{-120}
=0
$