What is the answer to (2(t^2-4))/(t^2+1)=1 ?

The answer to (2(t^2-4))/(t^2+1)=1 is t=3,t=-3

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(2(t^2-4))/(t^2+1)=1

\frac{2 ( t ^{2} - 4 )}{t ^{2} + 1} = 1

t = 3, t = - 3

Solution steps

\frac{2 ( t ^{2} - 4 )}{t ^{2} + 1} = 1

Multiply both sides by

t^{2} + 1

2 (t^{2} - 4) = t^{2} + 1

Solve

2 (t^{2} - 4) = t^{2} + 1 : t = 3, t = - 3

t = 3, t = - 3

a, 1 = \frac{7}{a} + \frac{11}{b} + \frac{24}{ab}

\frac{5}{x} = \frac{3}{x - 2}

\frac{81}{4} - 27 x + 9 x^{2} = \frac{81}{2 x}

\frac{2 - n - 3 n ^{2}}{3 n - 2} = n + 2

\frac{1800 x + 1440}{x + 1} = 1760

What is the answer to (2(t^2-4))/(t^2+1)=1 ?
The answer to (2(t^2-4))/(t^2+1)=1 is t=3,t=-3
Solve for t: (2(t^2-4))/(t^2+1)=1
The solution is t=3,t=-3