1*1/2z-2=3*1/4z-9=

Solution for 1*1/2z-2=3*1/4z-9= equation:

1*1/2z-2=3*1/4z-9=

We move all terms to the left:

1*1/2z-2-(3*1/4z-9)=0


Domain of the equation: 2z!=0

z!=0/2

z!=0

z∈R



Domain of the equation: 4z-9)!=0

z∈R


We get rid of parentheses

1*1/2z-3*1/4z+9-2=0

We calculate fractions


16z/8z^2+(-12z)/8z^2+9-2=0

We add all the numbers together, and all the variables

16z/8z^2+(-12z)/8z^2+7=0

We multiply all the terms by the denominator


16z+(-12z)+7*8z^2=0

Wy multiply elements

56z^2+16z+(-12z)=0

We get rid of parentheses

56z^2+16z-12z=0

We add all the numbers together, and all the variables

56z^2+4z=0

a = 56; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·56·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4}{2*56}=\frac{-8}{112}
=-1/14
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*56}=\frac{0}{112}
=0
$