What are the solutions to the equation -t^2+3t+27=0 ?

The solutions to the equation -t^2+3t+27=0 are t=-(-3+3sqrt(13))/2 ,t=(3+3sqrt(13))/2

Find the zeros of -t^2+3t+27=0

The zeros of -t^2+3t+27=0 are t=-(-3+3sqrt(13))/2 ,t=(3+3sqrt(13))/2

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- t^{2} + 3 t + 27 = 0

t = - \frac{- 3 + 3 13}{2}, t = \frac{3 + 3 13}{2}

Decimal Notation

t = - 3.90832 \dots, t = 6.90832 \dots

Solution steps

- t^{2} + 3 t + 27 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 ( - 1 ) \cdot 27}{2 ( - 1 )}

3^{2} - 4 (- 1) \cdot 27 = 313

t_{1, 2} = \frac{- 3 \pm 3 13}{2 ( - 1 )}

Separate the solutions

t_{1} = \frac{- 3 + 3 13}{2 ( - 1 )}, t_{2} = \frac{- 3 - 3 13}{2 ( - 1 )}

t = \frac{- 3 + 3 13}{2 ( - 1 )} : - \frac{- 3 + 3 13}{2}

t = \frac{- 3 - 3 13}{2 ( - 1 )} : \frac{3 + 3 13}{2}

The solutions to the quadratic equation are:

t = - \frac{- 3 + 3 13}{2}, t = \frac{3 + 3 13}{2}

x^{2} - 14 x + 49 = 0

2 x^{2} + 3 x + \frac{9}{8} = 0

2 (x^{2} + 9) = 26

- 10 x + 175 - 5 x^{2} = 0

192 x - 6 x^{2} = 0

What are the solutions to the equation -t^2+3t+27=0 ?
The solutions to the equation -t^2+3t+27=0 are t=-(-3+3sqrt(13))/2 ,t=(3+3sqrt(13))/2
Find the zeros of -t^2+3t+27=0
The zeros of -t^2+3t+27=0 are t=-(-3+3sqrt(13))/2 ,t=(3+3sqrt(13))/2