(t-4)^2/5-3=1

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Solution for (t-4)^2/5-3=1 equation:


t in (-oo:+oo)

((t-4)^2)/5-3 = 1 // - 1

((t-4)^2)/5-3-1 = 0

((t-4)^2)/5+(-3*5)/5+(-1*5)/5 = 0

(t-4)^2-3*5-1*5 = 0

t^2-8*t-5+1 = 0

t^2-8*t-4 = 0

t^2-8*t-4 = 0

t^2-8*t-4 = 0

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

DELTA = 80

DELTA > 0

t = (80^(1/2)+8)/(1*2) or t = (8-80^(1/2))/(1*2)

t = (4*5^(1/2)+8)/2 or t = (8-4*5^(1/2))/2

(t-((8-4*5^(1/2))/2))*(t-((4*5^(1/2)+8)/2)) = 0

((t-((8-4*5^(1/2))/2))*(t-((4*5^(1/2)+8)/2)))/5 = 0

((t-((8-4*5^(1/2))/2))*(t-((4*5^(1/2)+8)/2)))/5 = 0 // * 5

(t-((8-4*5^(1/2))/2))*(t-((4*5^(1/2)+8)/2)) = 0

( t-((4*5^(1/2)+8)/2) )

t-((4*5^(1/2)+8)/2) = 0 // + (4*5^(1/2)+8)/2

t = (4*5^(1/2)+8)/2

( t-((8-4*5^(1/2))/2) )

t-((8-4*5^(1/2))/2) = 0 // + (8-4*5^(1/2))/2

t = (8-4*5^(1/2))/2

t in { (4*5^(1/2)+8)/2, (8-4*5^(1/2))/2 }

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