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