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1/5x+3=1/2x-12
We move all terms to the left:
1/5x+3-(1/2x-12)=0
Domain of the equation: 5x!=0
x!=0/5
x!=0
x∈R
Domain of the equation: 2x-12)!=0We get rid of parentheses
x∈R
1/5x-1/2x+12+3=0
We calculate fractions
2x/10x^2+(-5x)/10x^2+12+3=0
We add all the numbers together, and all the variables
2x/10x^2+(-5x)/10x^2+15=0
We multiply all the terms by the denominator
2x+(-5x)+15*10x^2=0
Wy multiply elements
150x^2+2x+(-5x)=0
We get rid of parentheses
150x^2+2x-5x=0
We add all the numbers together, and all the variables
150x^2-3x=0
a = 150; b = -3; c = 0;
Δ = b2-4ac
Δ = -32-4·150·0
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3}{2*150}=\frac{0}{300} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3}{2*150}=\frac{6}{300} =1/50 $
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