What are the solutions to the equation 26=54t-5t^2 ?

The solutions to the equation 26=54t-5t^2 are t=(27-sqrt(599))/5 ,t=(27+sqrt(599))/5

Find the zeros of 26=54t-5t^2

The zeros of 26=54t-5t^2 are t=(27-sqrt(599))/5 ,t=(27+sqrt(599))/5

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26 = 54 t - 5 t^{2}

t = \frac{27 - 599}{5}, t = \frac{27 + 599}{5}

Decimal Notation

t = 0.50510 \dots, t = 10.29489 \dots

Solution steps

26 = 54 t - 5 t^{2}

Switch sides

54 t - 5 t^{2} = 26

Move

26

to the left side

54 t - 5 t^{2} - 26 = 0

Write in the standard form

a x^{2} + b x + c = 0

- 5 t^{2} + 54 t - 26 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 54 \pm 5 4 ^{2} - 4 ( - 5 ) ( - 26 )}{2 ( - 5 )}

5 4^{2} - 4 (- 5) (- 26) = 2599

t_{1, 2} = \frac{- 54 \pm 2 599}{2 ( - 5 )}

Separate the solutions

t_{1} = \frac{- 54 + 2 599}{2 ( - 5 )}, t_{2} = \frac{- 54 - 2 599}{2 ( - 5 )}

t = \frac{- 54 + 2 599}{2 ( - 5 )} : \frac{27 - 599}{5}

t = \frac{- 54 - 2 599}{2 ( - 5 )} : \frac{27 + 599}{5}

The solutions to the quadratic equation are:

t = \frac{27 - 599}{5}, t = \frac{27 + 599}{5}

t^{2} + t + 1 = (t + 1)^{2}

25 t^{2} - 120 t + 144

lo g_{\frac{1}{4}} (1024)

8 (2 x - 1) - \frac{1}{4} x = \frac{7}{4} (3 - 2 x)

\frac{7}{8} - 1

What are the solutions to the equation 26=54t-5t^2 ?
The solutions to the equation 26=54t-5t^2 are t=(27-sqrt(599))/5 ,t=(27+sqrt(599))/5
Find the zeros of 26=54t-5t^2
The zeros of 26=54t-5t^2 are t=(27-sqrt(599))/5 ,t=(27+sqrt(599))/5