1z+1=1/2z+4

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

1z+1=1/2z+4

We move all terms to the left:

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


Domain of the equation: 2z+4)!=0

z∈R


We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We multiply all the terms by the denominator


z*2z-4*2z+1*2z-1=0

Wy multiply elements

2z^2-8z+2z-1=0

We add all the numbers together, and all the variables

2z^2-6z-1=0

a = 2; b = -6; c = -1;
Δ = b2-4ac
Δ = -62-4·2·(-1)
Δ = 44
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:

$\sqrt{\Delta}=\sqrt{44}=\sqrt{4*11}=\sqrt{4}*\sqrt{11}=2\sqrt{11}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{11}}{2*2}=\frac{6-2\sqrt{11}}{4}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{11}}{2*2}=\frac{6+2\sqrt{11}}{4}
$