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2/5x-3/10x=12
We move all terms to the left:
2/5x-3/10x-(12)=0
Domain of the equation: 5x!=0
x!=0/5
x!=0
x∈R
Domain of the equation: 10x!=0We calculate fractions
x!=0/10
x!=0
x∈R
20x/50x^2+(-15x)/50x^2-12=0
We multiply all the terms by the denominator
20x+(-15x)-12*50x^2=0
Wy multiply elements
-600x^2+20x+(-15x)=0
We get rid of parentheses
-600x^2+20x-15x=0
We add all the numbers together, and all the variables
-600x^2+5x=0
a = -600; b = 5; c = 0;
Δ = b2-4ac
Δ = 52-4·(-600)·0
Δ = 25
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{25}=5$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5}{2*-600}=\frac{-10}{-1200} =1/120 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5}{2*-600}=\frac{0}{-1200} =0 $
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