2/3z+9=34-3/4z

Solution for 2/3z+9=34-3/4z equation:

2/3z+9=34-3/4z

We move all terms to the left:

2/3z+9-(34-3/4z)=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

2/3z-(-3/4z+34)+9=0

We get rid of parentheses

2/3z+3/4z-34+9=0

We calculate fractions


8z/12z^2+9z/12z^2-34+9=0

We add all the numbers together, and all the variables

8z/12z^2+9z/12z^2-25=0

We multiply all the terms by the denominator


8z+9z-25*12z^2=0

We add all the numbers together, and all the variables

17z-25*12z^2=0

Wy multiply elements

-300z^2+17z=0

a = -300; b = 17; c = 0;
Δ = b2-4ac
Δ = 172-4·(-300)·0
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-17}{2*-300}=\frac{-34}{-600}
=17/300
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+17}{2*-300}=\frac{0}{-600}
=0
$