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3/2x+4=2/5x+3
We move all terms to the left:
3/2x+4-(2/5x+3)=0
Domain of the equation: 2x!=0
x!=0/2
x!=0
x∈R
Domain of the equation: 5x+3)!=0We get rid of parentheses
x∈R
3/2x-2/5x-3+4=0
We calculate fractions
15x/10x^2+(-4x)/10x^2-3+4=0
We add all the numbers together, and all the variables
15x/10x^2+(-4x)/10x^2+1=0
We multiply all the terms by the denominator
15x+(-4x)+1*10x^2=0
Wy multiply elements
10x^2+15x+(-4x)=0
We get rid of parentheses
10x^2+15x-4x=0
We add all the numbers together, and all the variables
10x^2+11x=0
a = 10; b = 11; c = 0;
Δ = b2-4ac
Δ = 112-4·10·0
Δ = 121
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{121}=11$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-11}{2*10}=\frac{-22}{20} =-1+1/10 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+11}{2*10}=\frac{0}{20} =0 $
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