What are the solutions to the equation 2t+(t-4)^2=5t(t-4)+24 ?

The solutions to the equation 2t+(t-4)^2=5t(t-4)+24 are t=(7-sqrt(17))/4 ,t=(7+sqrt(17))/4

Find the zeros of 2t+(t-4)^2=5t(t-4)+24

The zeros of 2t+(t-4)^2=5t(t-4)+24 are t=(7-sqrt(17))/4 ,t=(7+sqrt(17))/4

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2 t + (t - 4)^{2} = 5 t (t - 4) + 24

t = \frac{7 - 17}{4}, t = \frac{7 + 17}{4}

Decimal Notation

t = 0.71922 \dots, t = 2.78077 \dots

Solution steps

2 t + (t - 4)^{2} = 5 t (t - 4) + 24

Expand

2 t + (t - 4)^{2} : t^{2} - 6 t + 16

Expand

5 t (t - 4) + 24 : 5 t^{2} - 20 t + 24

t^{2} - 6 t + 16 = 5 t^{2} - 20 t + 24

Move

24

to the left side

t^{2} - 6 t - 8 = 5 t^{2} - 20 t

Move

20 t

to the left side

t^{2} + 14 t - 8 = 5 t^{2}

Move

5 t^{2}

to the left side

- 4 t^{2} + 14 t - 8 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 14 \pm 1 4 ^{2} - 4 ( - 4 ) ( - 8 )}{2 ( - 4 )}

1 4^{2} - 4 (- 4) (- 8) = 217

t_{1, 2} = \frac{- 14 \pm 2 17}{2 ( - 4 )}

Separate the solutions

t_{1} = \frac{- 14 + 2 17}{2 ( - 4 )}, t_{2} = \frac{- 14 - 2 17}{2 ( - 4 )}

t = \frac{- 14 + 2 17}{2 ( - 4 )} : \frac{7 - 17}{4}

t = \frac{- 14 - 2 17}{2 ( - 4 )} : \frac{7 + 17}{4}

The solutions to the quadratic equation are:

t = \frac{7 - 17}{4}, t = \frac{7 + 17}{4}

(\frac{3}{5})^{- 4}

\frac{3 ^{- 8}}{3 ^{- 4}}

6 k^{2} + 2 k + 9 = - 3

x^{2} - 4 x = 52

- 4 u < - 16

What are the solutions to the equation 2t+(t-4)^2=5t(t-4)+24 ?
The solutions to the equation 2t+(t-4)^2=5t(t-4)+24 are t=(7-sqrt(17))/4 ,t=(7+sqrt(17))/4
Find the zeros of 2t+(t-4)^2=5t(t-4)+24
The zeros of 2t+(t-4)^2=5t(t-4)+24 are t=(7-sqrt(17))/4 ,t=(7+sqrt(17))/4