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24/5z=4/z-1
We move all terms to the left:
24/5z-(4/z-1)=0
Domain of the equation: 5z!=0
z!=0/5
z!=0
z∈R
Domain of the equation: z-1)!=0We get rid of parentheses
z∈R
24/5z-4/z+1=0
We calculate fractions
24z/5z^2+(-20z)/5z^2+1=0
We multiply all the terms by the denominator
24z+(-20z)+1*5z^2=0
Wy multiply elements
5z^2+24z+(-20z)=0
We get rid of parentheses
5z^2+24z-20z=0
We add all the numbers together, and all the variables
5z^2+4z=0
a = 5; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·5·0
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$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4}{2*5}=\frac{-8}{10} =-4/5 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*5}=\frac{0}{10} =0 $
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