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

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

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

We move all terms to the left:

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


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

z∈R


We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We multiply all the terms by the denominator


(+z^2+1-4*2*2)=0

We get rid of parentheses

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

We add all the numbers together, and all the variables

z^2-15=0

a = 1; b = 0; c = -15;
Δ = b2-4ac
Δ = 02-4·1·(-15)
Δ = 60
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:

$\sqrt{\Delta}=\sqrt{60}=\sqrt{4*15}=\sqrt{4}*\sqrt{15}=2\sqrt{15}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{15}}{2*1}=\frac{0-2\sqrt{15}}{2}
=-\frac{2\sqrt{15}}{2}
=-\sqrt{15}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{15}}{2*1}=\frac{0+2\sqrt{15}}{2}
=\frac{2\sqrt{15}}{2}
=\sqrt{15}
$