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3x=2+(3x/2)
We move all terms to the left:
3x-(2+(3x/2))=0
We add all the numbers together, and all the variables
3x-(2+(+3x/2))=0
We multiply all the terms by the denominator
3x*2))-(2+(+3x=0
We add all the numbers together, and all the variables
3x+3x*2))-(2+(=0
Wy multiply elements
6x^2+3x=0
a = 6; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·6·0
Δ = 9
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{9}=3$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*6}=\frac{-6}{12} =-1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*6}=\frac{0}{12} =0 $
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