(3x-1)^2/5=2

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Solution for (3x-1)^2/5=2 equation: 


x in (-oo:+oo)

((3*x-1)^2)/5 = 2 // - 2

((3*x-1)^2)/5-2 = 0

((3*x-1)^2)/5+(-2*5)/5 = 0

(3*x-1)^2-2*5 = 0

9*x^2-6*x-9 = 0

9*x^2-6*x-9 = 0

3*(3*x^2-2*x-3) = 0

3*x^2-2*x-3 = 0

DELTA = (-2)^2-(-3*3*4)

DELTA = 40

DELTA > 0

x = (40^(1/2)+2)/(2*3) or x = (2-40^(1/2))/(2*3)

x = (2*10^(1/2)+2)/6 or x = (2-2*10^(1/2))/6

3*(x-((2-2*10^(1/2))/6))*(x-((2*10^(1/2)+2)/6)) = 0

(3*(x-((2-2*10^(1/2))/6))*(x-((2*10^(1/2)+2)/6)))/5 = 0

(3*(x-((2-2*10^(1/2))/6))*(x-((2*10^(1/2)+2)/6)))/5 = 0 // * 5

3*(x-((2-2*10^(1/2))/6))*(x-((2*10^(1/2)+2)/6)) = 0

( 3 )

3 = 0

x belongs to the empty set

( x-((2*10^(1/2)+2)/6) )

x-((2*10^(1/2)+2)/6) = 0 // + (2*10^(1/2)+2)/6

x = (2*10^(1/2)+2)/6

( x-((2-2*10^(1/2))/6) )

x-((2-2*10^(1/2))/6) = 0 // + (2-2*10^(1/2))/6

x = (2-2*10^(1/2))/6

x in { (2*10^(1/2)+2)/6, (2-2*10^(1/2))/6 }

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