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

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


x in (-oo:+oo)

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

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

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

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

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

4*x^2-15*x+9 = 0

4*x^2-15*x+9 = 0

4*x^2-15*x+9 = 0

DELTA = (-15)^2-(4*4*9)

DELTA = 81

DELTA > 0

x = (81^(1/2)+15)/(2*4) or x = (15-81^(1/2))/(2*4)

x = 3 or x = 3/4

(x-3/4)*(x-3) = 0

((x-3/4)*(x-3))/5 = 0

((x-3/4)*(x-3))/5 = 0 // * 5

(x-3/4)*(x-3) = 0

( x-3/4 )

x-3/4 = 0 // + 3/4

x = 3/4

( x-3 )

x-3 = 0 // + 3

x = 3

x in { 3/4, 3 }

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