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1/2z-1/4z=2-5
We move all terms to the left:
1/2z-1/4z-(2-5)=0
Domain of the equation: 2z!=0
z!=0/2
z!=0
z∈R
Domain of the equation: 4z!=0We add all the numbers together, and all the variables
z!=0/4
z!=0
z∈R
1/2z-1/4z-(-3)=0
We add all the numbers together, and all the variables
1/2z-1/4z+3=0
We calculate fractions
4z/8z^2+(-2z)/8z^2+3=0
We multiply all the terms by the denominator
4z+(-2z)+3*8z^2=0
Wy multiply elements
24z^2+4z+(-2z)=0
We get rid of parentheses
24z^2+4z-2z=0
We add all the numbers together, and all the variables
24z^2+2z=0
a = 24; b = 2; c = 0;
Δ = b2-4ac
Δ = 22-4·24·0
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$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2}{2*24}=\frac{-4}{48} =-1/12 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2}{2*24}=\frac{0}{48} =0 $
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