1/5(x-4)^2=21

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Solution for 1/5(x-4)^2=21 equation: 


x in (-oo:+oo)

(1/5)*(x-4)^2 = 21 // - 21

(1/5)*(x-4)^2-21 = 0

1/5*(x-4)^2-21 = 0

1/5*(x-4)^2-21 = 0

1/5*x^2-8/5*x-89/5 = 0

1/5*x^2-8/5*x-89/5 = 0

1/5*(x^2-8*x-89) = 0

x^2-8*x-89 = 0

DELTA = (-8)^2-(-89*1*4)

DELTA = 420

DELTA > 0

x = (420^(1/2)+8)/(1*2) or x = (8-420^(1/2))/(1*2)

x = (2*105^(1/2)+8)/2 or x = (8-2*105^(1/2))/2

1/5*(x-((8-2*105^(1/2))/2))*(x-((2*105^(1/2)+8)/2)) = 0

1/5*(x-((8-2*105^(1/2))/2))*(x-((2*105^(1/2)+8)/2)) = 0

( 1/5 )

1/5 = 0

x belongs to the empty set

( x-((2*105^(1/2)+8)/2) )

x-((2*105^(1/2)+8)/2) = 0 // + (2*105^(1/2)+8)/2

x = (2*105^(1/2)+8)/2

( x-((8-2*105^(1/2))/2) )

x-((8-2*105^(1/2))/2) = 0 // + (8-2*105^(1/2))/2

x = (8-2*105^(1/2))/2

x in { (2*105^(1/2)+8)/2, (8-2*105^(1/2))/2 }

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