10(x-5)2/5-25=15

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Solution for 10(x-5)2/5-25=15 equation: 


x in (-oo:+oo)

(10*(x-5)^2)/5-25 = 15 // - 15

(10*(x-5)^2)/5-25-15 = 0

(10*(x-5)^2)/5+(-25*5)/5+(-15*5)/5 = 0

10*(x-5)^2-25*5-15*5 = 0

10*x^2-100*x-75+125 = 0

10*x^2-100*x+50 = 0

10*x^2-100*x+50 = 0

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

x^2-10*x+5 = 0

DELTA = (-10)^2-(1*4*5)

DELTA = 80

DELTA > 0

x = (80^(1/2)+10)/(1*2) or x = (10-80^(1/2))/(1*2)

x = (4*5^(1/2)+10)/2 or x = (10-4*5^(1/2))/2

10*(x-((10-4*5^(1/2))/2))*(x-((4*5^(1/2)+10)/2)) = 0

(10*(x-((10-4*5^(1/2))/2))*(x-((4*5^(1/2)+10)/2)))/5 = 0

(10*(x-((10-4*5^(1/2))/2))*(x-((4*5^(1/2)+10)/2)))/5 = 0 // * 5

10*(x-((10-4*5^(1/2))/2))*(x-((4*5^(1/2)+10)/2)) = 0

( 10 )

10 = 0

x belongs to the empty set

( x-((10-4*5^(1/2))/2) )

x-((10-4*5^(1/2))/2) = 0 // + (10-4*5^(1/2))/2

x = (10-4*5^(1/2))/2

( x-((4*5^(1/2)+10)/2) )

x-((4*5^(1/2)+10)/2) = 0 // + (4*5^(1/2)+10)/2

x = (4*5^(1/2)+10)/2

x in { (10-4*5^(1/2))/2, (4*5^(1/2)+10)/2 }

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