1/5(x-4)^2=5

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Solution for 1/5(x-4)^2=5 equation: 


x in (-oo:+oo)

(1/5)*(x-4)^2 = 5 // - 5

(1/5)*(x-4)^2-5 = 0

1/5*(x-4)^2-5 = 0

1/5*(x-4)^2-5 = 0

1/5*x^2-8/5*x-9/5 = 0

1/5*x^2-8/5*x-9/5 = 0

1/5*(x^2-8*x-9) = 0

x^2-8*x-9 = 0

DELTA = (-8)^2-(-9*1*4)

DELTA = 100

DELTA > 0

x = (100^(1/2)+8)/(1*2) or x = (8-100^(1/2))/(1*2)

x = 9 or x = -1

1/5*(x+1)*(x-9) = 0

1/5*(x+1)*(x-9) = 0

( 1/5 )

1/5 = 0

x belongs to the empty set

( x+1 )

x+1 = 0 // - 1

x = -1

( x-9 )

x-9 = 0 // + 9

x = 9

x in { -1, 9 }

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