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3x^+4x/2x+9=5
We move all terms to the left:
3x^+4x/2x+9-(5)=0
Domain of the equation: 2x!=0We add all the numbers together, and all the variables
x!=0/2
x!=0
x∈R
3x+4x/2x+4=0
We multiply all the terms by the denominator
3x*2x+4x+4*2x=0
We add all the numbers together, and all the variables
4x+3x*2x+4*2x=0
Wy multiply elements
6x^2+4x+8x=0
We add all the numbers together, and all the variables
6x^2+12x=0
a = 6; b = 12; c = 0;
Δ = b2-4ac
Δ = 122-4·6·0
Δ = 144
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{144}=12$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-12}{2*6}=\frac{-24}{12} =-2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12}{2*6}=\frac{0}{12} =0 $
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