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

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

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

We move all terms to the left:

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

We get rid of parentheses

-7z+1/3z-8+12=0

We multiply all the terms by the denominator


-7z*3z-8*3z+12*3z+1=0

Wy multiply elements

-21z^2-24z+36z+1=0

We add all the numbers together, and all the variables

-21z^2+12z+1=0

a = -21; b = 12; c = +1;
Δ = b2-4ac
Δ = 122-4·(-21)·1
Δ = 228
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:

$\sqrt{\Delta}=\sqrt{228}=\sqrt{4*57}=\sqrt{4}*\sqrt{57}=2\sqrt{57}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{57}}{2*-21}=\frac{-12-2\sqrt{57}}{-42}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{57}}{2*-21}=\frac{-12+2\sqrt{57}}{-42}
$