What are the solutions to the equation 1+t^2=2 ?

The solutions to the equation 1+t^2=2 are t=1,t=-1

Find the zeros of 1+t^2=2

The zeros of 1+t^2=2 are t=1,t=-1

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1 + t^{2} = 2

t = 1, t = - 1

Solution steps

1 + t^{2} = 2

Move

1

to the right side

t^{2} = 1

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

t = 1, t = - 1

1 = 1

- 1 = - 1

t = 1, t = - 1

x^{2} - 19 x - 39 = 0

x, x^{2} y + 3 x - 6 y = 4

4 x^{2} + 0.76 x + 1.96 = 0

\frac{x ^{2}}{15} - \frac{x}{5} - \frac{2}{3} = 0

\frac{3}{5} x^{2} + 3 x + 1 = - 2 - 3 x

What are the solutions to the equation 1+t^2=2 ?
The solutions to the equation 1+t^2=2 are t=1,t=-1
Find the zeros of 1+t^2=2
The zeros of 1+t^2=2 are t=1,t=-1