What is the answer to (t^2+1)/(t^2+2t-1)=0 ?

The answer to (t^2+1)/(t^2+2t-1)=0 is t=i,t=-i

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(t^2+1)/(t^2+2t-1)=0

\frac{t ^{2} + 1}{t ^{2} + 2 t - 1} = 0

t = i, t = - i

Solution steps

\frac{t ^{2} + 1}{t ^{2} + 2 t - 1} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

t^{2} + 1 = 0

Solve

t^{2} + 1 = 0 : t = i, t = - i

t = i, t = - i

\frac{9}{x + 1} = \frac{18}{54}

\frac{9}{x} = \frac{15}{x - 2}

\frac{7}{x ^{2} - 1} = \frac{5}{2 x}

\frac{6 x}{x + 4} = \frac{5 x}{x + 3}

A, 0.8 = \frac{3680}{P A}

What is the answer to (t^2+1)/(t^2+2t-1)=0 ?
The answer to (t^2+1)/(t^2+2t-1)=0 is t=i,t=-i
Solve for t: (t^2+1)/(t^2+2t-1)=0
The solution is t=i,t=-i