What is the answer to c=(20t)/(t^2+4) ?

The answer to c=(20t)/(t^2+4) is t=(2(sqrt(-c^2+25)+5))/c ,t=(2(-sqrt(-c^2+25)+5))/c ;\quad c\ne 0

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c=(20t)/(t^2+4)

c = \frac{20 t}{t ^{2} + 4}

t = \frac{2 ( - c ^{2} + 25 + 5 )}{c}, t = \frac{2 ( - - c ^{2} + 25 + 5 )}{c}; c \neq = 0

Solution steps

c = \frac{20 t}{t ^{2} + 4}

Multiply both sides by

t^{2} + 4

c (t^{2} + 4) = 20 t

Solve

c (t^{2} + 4) = 20 t : t = \frac{2 ( - c ^{2} + 25 + 5 )}{c}, t = \frac{2 ( - - c ^{2} + 25 + 5 )}{c}; c \neq = 0

t = \frac{2 ( - c ^{2} + 25 + 5 )}{c}, t = \frac{2 ( - - c ^{2} + 25 + 5 )}{c}; c \neq = 0

4 x^{2} + x^{- 1} - 3 = 0

\frac{x}{x + 4} + \frac{4}{x + 4} = - 2

4 - \frac{8000}{x ^{2}} = 0

\frac{2 x}{3} + \frac{3}{2 x} = 1

\frac{27}{x} + 41 = 50

What is the answer to c=(20t)/(t^2+4) ?
The answer to c=(20t)/(t^2+4) is t=(2(sqrt(-c^2+25)+5))/c ,t=(2(-sqrt(-c^2+25)+5))/c ;\quad c\ne 0