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

Solution for 1+1/2*z-2=3+1/4*z-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 add all the numbers together, and all the variables

1/2z-(1/4z-6)+1-2=0

We add all the numbers together, and all the variables

1/2z-(1/4z-6)-1=0

We get rid of parentheses

1/2z-1/4z+6-1=0

We calculate fractions


4z/8z^2+(-2z)/8z^2+6-1=0

We add all the numbers together, and all the variables

4z/8z^2+(-2z)/8z^2+5=0

We multiply all the terms by the denominator


4z+(-2z)+5*8z^2=0

Wy multiply elements

40z^2+4z+(-2z)=0

We get rid of parentheses

40z^2+4z-2z=0

We add all the numbers together, and all the variables

40z^2+2z=0

a = 40; b = 2; c = 0;
Δ = b2-4ac
Δ = 22-4·40·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2}{2*40}=\frac{-4}{80}
=-1/20
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2}{2*40}=\frac{0}{80}
=0
$