What are the solutions to the equation -t^2+2t-1=0 ?

The solutions to the equation -t^2+2t-1=0 are t=1

Find the zeros of -t^2+2t-1=0

The zeros of -t^2+2t-1=0 are t=1

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

t = 1

Solution steps

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

Solve with the quadratic formula

t_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 ( - 1 ) ( - 1 )}{2 ( - 1 )}

2^{2} - 4 (- 1) (- 1) = 0

t_{1, 2} = \frac{- 2 \pm 0}{2 ( - 1 )}

t = \frac{- 2}{2 ( - 1 )}

\frac{- 2}{2 ( - 1 )} = 1

t = 1

The solution to the quadratic equation is:

t = 1

(a + 1) x^{2} + (a - 2) x + 2 a + 3 = 0

x^{2} - 7 x + 10 = 4

x^{2} = 48 - 2 x

2 x^{2} - 35 = 1

4 x^{2} + 21 = 5 x

What are the solutions to the equation -t^2+2t-1=0 ?
The solutions to the equation -t^2+2t-1=0 are t=1
Find the zeros of -t^2+2t-1=0
The zeros of -t^2+2t-1=0 are t=1