What are the solutions to the equation 0=-t^2+4t+5 ?

The solutions to the equation 0=-t^2+4t+5 are t=-1,t=5

Find the zeros of 0=-t^2+4t+5

The zeros of 0=-t^2+4t+5 are t=-1,t=5

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0 = - t^{2} + 4 t + 5

t = - 1, t = 5

Solution steps

0 = - t^{2} + 4 t + 5

Switch sides

- t^{2} + 4 t + 5 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 ( - 1 ) \cdot 5}{2 ( - 1 )}

4^{2} - 4 (- 1) \cdot 5 = 6

t_{1, 2} = \frac{- 4 \pm 6}{2 ( - 1 )}

Separate the solutions

t_{1} = \frac{- 4 + 6}{2 ( - 1 )}, t_{2} = \frac{- 4 - 6}{2 ( - 1 )}

t = \frac{- 4 + 6}{2 ( - 1 )} : - 1

t = \frac{- 4 - 6}{2 ( - 1 )} : 5

The solutions to the quadratic equation are:

t = - 1, t = 5

9 ∣ m - 8 ∣ - 10 < 26

v^{2} + 6 v - 7 = 0

∣ b ∣ = 6

5 < 3 x - 1 \leq 14

5 r - 2 = 6 r + 5

What are the solutions to the equation 0=-t^2+4t+5 ?
The solutions to the equation 0=-t^2+4t+5 are t=-1,t=5
Find the zeros of 0=-t^2+4t+5
The zeros of 0=-t^2+4t+5 are t=-1,t=5