What are the solutions to the equation 28=-t^2+3t+27 ?

The solutions to the equation 28=-t^2+3t+27 are t=(3-sqrt(5))/2 ,t=(3+sqrt(5))/2

Find the zeros of 28=-t^2+3t+27

The zeros of 28=-t^2+3t+27 are t=(3-sqrt(5))/2 ,t=(3+sqrt(5))/2

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

t = \frac{3 - 5}{2}, t = \frac{3 + 5}{2}

Solution steps

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

Switch sides

- t^{2} + 3 t + 27 = 28

Move

28

to the left side

- t^{2} + 3 t - 1 = 0

Solve with the quadratic formula

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

3^{2} - 4 (- 1) (- 1) = 5

t_{1, 2} = \frac{- 3 \pm 5}{2 ( - 1 )}

Separate the solutions

t_{1} = \frac{- 3 + 5}{2 ( - 1 )}, t_{2} = \frac{- 3 - 5}{2 ( - 1 )}

t = \frac{- 3 + 5}{2 ( - 1 )} : \frac{3 - 5}{2}

t = \frac{- 3 - 5}{2 ( - 1 )} : \frac{3 + 5}{2}

The solutions to the quadratic equation are:

t = \frac{3 - 5}{2}, t = \frac{3 + 5}{2}

x^{2} - 8 x + 16 = 36

4^{2} + b^{2} = 7^{2}

1 0^{2} + y^{2} = 1 2^{2}

p^{2} + 8 p + 15 = 0

3 x^{2} - 8 x + 12 = 0

What are the solutions to the equation 28=-t^2+3t+27 ?
The solutions to the equation 28=-t^2+3t+27 are t=(3-sqrt(5))/2 ,t=(3+sqrt(5))/2
Find the zeros of 28=-t^2+3t+27
The zeros of 28=-t^2+3t+27 are t=(3-sqrt(5))/2 ,t=(3+sqrt(5))/2