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2/3x+2=1/2x-5
We move all terms to the left:
2/3x+2-(1/2x-5)=0
Domain of the equation: 3x!=0
x!=0/3
x!=0
x∈R
Domain of the equation: 2x-5)!=0We get rid of parentheses
x∈R
2/3x-1/2x+5+2=0
We calculate fractions
4x/6x^2+(-3x)/6x^2+5+2=0
We add all the numbers together, and all the variables
4x/6x^2+(-3x)/6x^2+7=0
We multiply all the terms by the denominator
4x+(-3x)+7*6x^2=0
Wy multiply elements
42x^2+4x+(-3x)=0
We get rid of parentheses
42x^2+4x-3x=0
We add all the numbers together, and all the variables
42x^2+x=0
a = 42; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·42·0
Δ = 1
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{1}=1$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*42}=\frac{-2}{84} =-1/42 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*42}=\frac{0}{84} =0 $
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