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x+8/2x-5=1
We move all terms to the left:
x+8/2x-5-(1)=0
Domain of the equation: 2x!=0We add all the numbers together, and all the variables
x!=0/2
x!=0
x∈R
x+8/2x-6=0
We multiply all the terms by the denominator
x*2x-6*2x+8=0
Wy multiply elements
2x^2-12x+8=0
a = 2; b = -12; c = +8;
Δ = b2-4ac
Δ = -122-4·2·8
Δ = 80
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{5}}{2*2}=\frac{12-4\sqrt{5}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{5}}{2*2}=\frac{12+4\sqrt{5}}{4} $
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