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2x+3/2x-3=3
We move all terms to the left:
2x+3/2x-3-(3)=0
Domain of the equation: 2x!=0We add all the numbers together, and all the variables
x!=0/2
x!=0
x∈R
2x+3/2x-6=0
We multiply all the terms by the denominator
2x*2x-6*2x+3=0
Wy multiply elements
4x^2-12x+3=0
a = 4; b = -12; c = +3;
Δ = b2-4ac
Δ = -122-4·4·3
Δ = 96
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{6}}{2*4}=\frac{12-4\sqrt{6}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{6}}{2*4}=\frac{12+4\sqrt{6}}{8} $
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