3z+-1/2z+4=1

Solution for 3z+-1/2z+4=1 equation:

3z+-1/2z+4=1

We move all terms to the left:

3z+-1/2z+4-(1)=0


Domain of the equation: 2z!=0

z!=0/2

z!=0

z∈R


determiningTheFunctionDomain
3z-1/2z+4-1+=0

We add all the numbers together, and all the variables

3z-1/2z=0

We multiply all the terms by the denominator


3z*2z-1=0

Wy multiply elements

6z^2-1=0

a = 6; b = 0; c = -1;
Δ = b2-4ac
Δ = 02-4·6·(-1)
Δ = 24
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:

$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{6}}{2*6}=\frac{0-2\sqrt{6}}{12}
=-\frac{2\sqrt{6}}{12}
=-\frac{\sqrt{6}}{6}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{6}}{2*6}=\frac{0+2\sqrt{6}}{12}
=\frac{2\sqrt{6}}{12}
=\frac{\sqrt{6}}{6}
$