25=5t^(1/2)-0.25t

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Solution for 25=5t^(1/2)-0.25t equation: 


D( t )

t < 0

t < 0

t in <0:+oo)

25 = 5*t^(1/2)-(0.25*t) // - 5*t^(1/2)-(0.25*t)

0.25*t-(5*t^(1/2))+25 = 0

0.25*t-5*t^(1/2)+25 = 0

t_1 = t^(1/2)

0.25*t_1^2-5*t_1^1+25 = 0

0.25*t_1^2-5*t_1+25 = 0

DELTA = (-5)^2-(0.25*4*25)

DELTA = 0

t_1 = 5/(0.25*2)

t_1 = 10 or t_1 = 10

t_1 = 10

t^(1/2)-10 = 0

1*t^(1/2) = 10 // : 1

t^(1/2) = 10

t^(1/2) = 10 // ^ 2

t = 100

t_1 = 10

t^(1/2)-10 = 0

1*t^(1/2) = 10 // : 1

t^(1/2) = 10

t^(1/2) = 10 // ^ 2

t = 100

t in { 100, 100 }

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