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5z-(2z-9)=1/2z+7
We move all terms to the left:
5z-(2z-9)-(1/2z+7)=0
Domain of the equation: 2z+7)!=0We get rid of parentheses
z∈R
5z-2z-1/2z+9-7=0
We multiply all the terms by the denominator
5z*2z-2z*2z+9*2z-7*2z-1=0
Wy multiply elements
10z^2-4z^2+18z-14z-1=0
We add all the numbers together, and all the variables
6z^2+4z-1=0
a = 6; b = 4; c = -1;
Δ = b2-4ac
Δ = 42-4·6·(-1)
Δ = 40
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}$$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{10}}{2*6}=\frac{-4-2\sqrt{10}}{12} $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{10}}{2*6}=\frac{-4+2\sqrt{10}}{12} $
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