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

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

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

We move all terms to the left:

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


Domain of the equation: 2z!=0

z!=0/2

z!=0

z∈R


We add all the numbers together, and all the variables

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

We get rid of parentheses

1/2z+1/2z+3z-12+2=0

We multiply all the terms by the denominator


3z*2z-12*2z+2*2z+1+1=0

We add all the numbers together, and all the variables

3z*2z-12*2z+2*2z+2=0

Wy multiply elements

6z^2-24z+4z+2=0

We add all the numbers together, and all the variables

6z^2-20z+2=0

a = 6; b = -20; c = +2;
Δ = b2-4ac
Δ = -202-4·6·2
Δ = 352
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:

$\sqrt{\Delta}=\sqrt{352}=\sqrt{16*22}=\sqrt{16}*\sqrt{22}=4\sqrt{22}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{22}}{2*6}=\frac{20-4\sqrt{22}}{12}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{22}}{2*6}=\frac{20+4\sqrt{22}}{12}
$