-7z+12=2-z+8-1/3z

Solution for -7z+12=2-z+8-1/3z equation:

-7z+12=2-z+8-1/3z

We move all terms to the left:

-7z+12-(2-z+8-1/3z)=0


Domain of the equation: 3z)!=0

z!=0/1

z!=0

z∈R


We add all the numbers together, and all the variables

-7z-(-1z-1/3z+10)+12=0

We get rid of parentheses

-7z+1z+1/3z-10+12=0

We multiply all the terms by the denominator


-7z*3z+1z*3z-10*3z+12*3z+1=0

Wy multiply elements

-21z^2+3z^2-30z+36z+1=0

We add all the numbers together, and all the variables

-18z^2+6z+1=0

a = -18; b = 6; c = +1;
Δ = b2-4ac
Δ = 62-4·(-18)·1
Δ = 108
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:

$\sqrt{\Delta}=\sqrt{108}=\sqrt{36*3}=\sqrt{36}*\sqrt{3}=6\sqrt{3}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6\sqrt{3}}{2*-18}=\frac{-6-6\sqrt{3}}{-36}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6\sqrt{3}}{2*-18}=\frac{-6+6\sqrt{3}}{-36}
$