1/2z+3-z=-2

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

1/2z+3-z=-2

We move all terms to the left:

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


Domain of the equation: 2z!=0

z!=0/2

z!=0

z∈R


We add all the numbers together, and all the variables

-1z+1/2z+5=0

We multiply all the terms by the denominator


-1z*2z+5*2z+1=0

Wy multiply elements

-2z^2+10z+1=0

a = -2; b = 10; c = +1;
Δ = b2-4ac
Δ = 102-4·(-2)·1
Δ = 108
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:

$\sqrt{\Delta}=\sqrt{108}=\sqrt{36*3}=\sqrt{36}*\sqrt{3}=6\sqrt{3}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-6\sqrt{3}}{2*-2}=\frac{-10-6\sqrt{3}}{-4}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+6\sqrt{3}}{2*-2}=\frac{-10+6\sqrt{3}}{-4}
$