1/8(2x-5)^2=5

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Solution for 1/8(2x-5)^2=5 equation: 


x in (-oo:+oo)

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

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

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

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

1/2*x^2-5/2*x-15/8 = 0

1/2*x^2-5/2*x-15/8 = 0

1/2*(x^2-5*x-15/4) = 0

x^2-5*x-15/4 = 0

DELTA = (-5)^2-(-15/4*1*4)

DELTA = 40

DELTA > 0

x = (40^(1/2)+5)/(1*2) or x = (5-40^(1/2))/(1*2)

x = (2*10^(1/2)+5)/2 or x = (5-2*10^(1/2))/2

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

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

( 1/2 )

1/2 = 0

x belongs to the empty set

( x-((2*10^(1/2)+5)/2) )

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

x = (2*10^(1/2)+5)/2

( x-((5-2*10^(1/2))/2) )

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

x = (5-2*10^(1/2))/2

x in { (2*10^(1/2)+5)/2, (5-2*10^(1/2))/2 }

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