41/6z=-2+52/3z

Solution for 41/6z=-2+52/3z equation:

41/6z=-2+52/3z

We move all terms to the left:

41/6z-(-2+52/3z)=0


Domain of the equation: 6z!=0

z!=0/6

z!=0

z∈R



Domain of the equation: 3z)!=0

z!=0/1

z!=0

z∈R


We add all the numbers together, and all the variables

41/6z-(52/3z-2)=0

We get rid of parentheses

41/6z-52/3z+2=0

We calculate fractions


123z/18z^2+(-312z)/18z^2+2=0

We multiply all the terms by the denominator


123z+(-312z)+2*18z^2=0

Wy multiply elements

36z^2+123z+(-312z)=0

We get rid of parentheses

36z^2+123z-312z=0

We add all the numbers together, and all the variables

36z^2-189z=0

a = 36; b = -189; c = 0;
Δ = b2-4ac
Δ = -1892-4·36·0
Δ = 35721
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{35721}=189$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-189)-189}{2*36}=\frac{0}{72}
=0
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-189)+189}{2*36}=\frac{378}{72}
=5+1/4
$