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z-6/2z+1=10
We move all terms to the left:
z-6/2z+1-(10)=0
Domain of the equation: 2z!=0We add all the numbers together, and all the variables
z!=0/2
z!=0
z∈R
z-6/2z-9=0
We multiply all the terms by the denominator
z*2z-9*2z-6=0
Wy multiply elements
2z^2-18z-6=0
a = 2; b = -18; c = -6;
Δ = b2-4ac
Δ = -182-4·2·(-6)
Δ = 372
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}$$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{372}=\sqrt{4*93}=\sqrt{4}*\sqrt{93}=2\sqrt{93}$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{93}}{2*2}=\frac{18-2\sqrt{93}}{4} $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{93}}{2*2}=\frac{18+2\sqrt{93}}{4} $
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