7z+12=20z+8-1/3z

Solution for 7z+12=20z+8-1/3z equation:

7z+12=20z+8-1/3z

We move all terms to the left:

7z+12-(20z+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-(20z-1/3z+8)+12=0

We get rid of parentheses

7z-20z+1/3z-8+12=0

We multiply all the terms by the denominator


7z*3z-20z*3z-8*3z+12*3z+1=0

Wy multiply elements

21z^2-60z^2-24z+36z+1=0

We add all the numbers together, and all the variables

-39z^2+12z+1=0

a = -39; b = 12; c = +1;
Δ = b2-4ac
Δ = 122-4·(-39)·1
Δ = 300
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:

$\sqrt{\Delta}=\sqrt{300}=\sqrt{100*3}=\sqrt{100}*\sqrt{3}=10\sqrt{3}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-10\sqrt{3}}{2*-39}=\frac{-12-10\sqrt{3}}{-78}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+10\sqrt{3}}{2*-39}=\frac{-12+10\sqrt{3}}{-78}
$