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7/2z+5=3/z
We move all terms to the left:
7/2z+5-(3/z)=0
Domain of the equation: 2z!=0
z!=0/2
z!=0
z∈R
Domain of the equation: z)!=0We add all the numbers together, and all the variables
z!=0/1
z!=0
z∈R
7/2z-(+3/z)+5=0
We get rid of parentheses
7/2z-3/z+5=0
We calculate fractions
7z/2z^2+(-6z)/2z^2+5=0
We multiply all the terms by the denominator
7z+(-6z)+5*2z^2=0
Wy multiply elements
10z^2+7z+(-6z)=0
We get rid of parentheses
10z^2+7z-6z=0
We add all the numbers together, and all the variables
10z^2+z=0
a = 10; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·10·0
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$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*10}=\frac{-2}{20} =-1/10 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*10}=\frac{0}{20} =0 $
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