-2/3z+9=9z-8

Solution for -2/3z+9=9z-8 equation:

-2/3z+9=9z-8

We move all terms to the left:

-2/3z+9-(9z-8)=0


Domain of the equation: 3z!=0

z!=0/3

z!=0

z∈R


We get rid of parentheses

-2/3z-9z+8+9=0

We multiply all the terms by the denominator


-9z*3z+8*3z+9*3z-2=0

Wy multiply elements

-27z^2+24z+27z-2=0

We add all the numbers together, and all the variables

-27z^2+51z-2=0

a = -27; b = 51; c = -2;
Δ = b2-4ac
Δ = 512-4·(-27)·(-2)
Δ = 2385
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:

$\sqrt{\Delta}=\sqrt{2385}=\sqrt{9*265}=\sqrt{9}*\sqrt{265}=3\sqrt{265}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(51)-3\sqrt{265}}{2*-27}=\frac{-51-3\sqrt{265}}{-54}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(51)+3\sqrt{265}}{2*-27}=\frac{-51+3\sqrt{265}}{-54}
$