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3x^=3/2x+3
We move all terms to the left:
3x^-(3/2x+3)=0
Domain of the equation: 2x+3)!=0We add all the numbers together, and all the variables
x∈R
3x-(3/2x+3)=0
We get rid of parentheses
3x-3/2x-3=0
We multiply all the terms by the denominator
3x*2x-3*2x-3=0
Wy multiply elements
6x^2-6x-3=0
a = 6; b = -6; c = -3;
Δ = b2-4ac
Δ = -62-4·6·(-3)
Δ = 108
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{108}=\sqrt{36*3}=\sqrt{36}*\sqrt{3}=6\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-6\sqrt{3}}{2*6}=\frac{6-6\sqrt{3}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+6\sqrt{3}}{2*6}=\frac{6+6\sqrt{3}}{12} $
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