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2/z-5=0.5z^2+7
We move all terms to the left:
2/z-5-(0.5z^2+7)=0
Domain of the equation: z!=0We get rid of parentheses
z∈R
-0.5z^2+2/z-7-5=0
We multiply all the terms by the denominator
-(0.5z^2)*z-7*z-5*z+2=0
We add all the numbers together, and all the variables
-12z-(0.5z^2)*z+2=0
We multiply parentheses
-0z^2-12z+2=0
We add all the numbers together, and all the variables
-1z^2-12z+2=0
a = -1; b = -12; c = +2;
Δ = b2-4ac
Δ = -122-4·(-1)·2
Δ = 152
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}$$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{152}=\sqrt{4*38}=\sqrt{4}*\sqrt{38}=2\sqrt{38}$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-2\sqrt{38}}{2*-1}=\frac{12-2\sqrt{38}}{-2} $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+2\sqrt{38}}{2*-1}=\frac{12+2\sqrt{38}}{-2} $
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