19+z=5/9z+3

Solution for 19+z=5/9z+3 equation:

19+z=5/9z+3

We move all terms to the left:

19+z-(5/9z+3)=0


Domain of the equation: 9z+3)!=0

z∈R


We get rid of parentheses

z-5/9z-3+19=0

We multiply all the terms by the denominator


z*9z-3*9z+19*9z-5=0

Wy multiply elements

9z^2-27z+171z-5=0

We add all the numbers together, and all the variables

9z^2+144z-5=0

a = 9; b = 144; c = -5;
Δ = b2-4ac
Δ = 1442-4·9·(-5)
Δ = 20916
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{20916}=\sqrt{36*581}=\sqrt{36}*\sqrt{581}=6\sqrt{581}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(144)-6\sqrt{581}}{2*9}=\frac{-144-6\sqrt{581}}{18}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(144)+6\sqrt{581}}{2*9}=\frac{-144+6\sqrt{581}}{18}
$