What are the solutions to the equation -t^2+2t+3=0 ?

The solutions to the equation -t^2+2t+3=0 are t=-1,t=3

Find the zeros of -t^2+2t+3=0

The zeros of -t^2+2t+3=0 are t=-1,t=3

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- t^{2} + 2 t + 3 = 0

t = - 1, t = 3

Solution steps

- t^{2} + 2 t + 3 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 ( - 1 ) \cdot 3}{2 ( - 1 )}

2^{2} - 4 (- 1) \cdot 3 = 4

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

Separate the solutions

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

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

t = \frac{- 2 - 4}{2 ( - 1 )} : 3

The solutions to the quadratic equation are:

t = - 1, t = 3

2 x^{2} - 6 x = - 1

k^{2} + 1 = 0

m^{2} + 45 m + 600 = 0

x^{2} + 3 b x - 10 b^{2} = 0

x^{2} = 8^{2} + 1 0^{2}

What are the solutions to the equation -t^2+2t+3=0 ?
The solutions to the equation -t^2+2t+3=0 are t=-1,t=3
Find the zeros of -t^2+2t+3=0
The zeros of -t^2+2t+3=0 are t=-1,t=3