z+2(z=3)+(z/2)=29

Solution for z+2(z=3)+(z/2)=29 equation:

z+2(z=3)+(z/2)=29

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply all the terms by the denominator


z*2))+2(z-3+(+z=0

We add all the numbers together, and all the variables

z+z*2))+2(z-3+(=0

Wy multiply elements

2z^2+z=0

a = 2; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·2·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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