What are the solutions to the equation -5t^2+(20+6e^{-4})t-(16+15e^{-4})=0 ?

The solutions to the equation -5t^2+(20+6e^{-4})t-(16+15e^{-4})=0 are

Find the zeros of -5t^2+(20+6e^{-4})t-(16+15e^{-4})=0

The zeros of -5t^2+(20+6e^{-4})t-(16+15e^{-4})=0 are

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- 5 t^{2} + (20 + 6 e^{- 4}) t - (16 + 15 e^{- 4}) = 0

t = \frac{6 + 20 e ^{4} - e ^{4} 80 + \frac{36}{e ^{8}} - \frac{60}{e ^{4}}}{10 e ^{4}}, t = \frac{6 + 20 e ^{4} + e ^{4} 80 + \frac{36}{e ^{8}} - \frac{60}{e ^{4}}}{10 e ^{4}}

Decimal Notation

t = 1.12265 \dots, t = 2.89932 \dots

Solution steps

- 5 t^{2} + (20 + 6 e^{- 4}) t - (16 + 15 e^{- 4}) = 0

Expand

- 5 t^{2} + (20 + 6 e^{- 4}) t - (16 + 15 e^{- 4}) : - 5 t^{2} + \frac{20 e ^{4} + 6}{e ^{4}} t - 16 - \frac{15}{e ^{4}}

- 5 t^{2} + \frac{20 e ^{4} + 6}{e ^{4}} t - 16 - \frac{15}{e ^{4}} = 0

Write in the standard form

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

- 5 t^{2} + (20 + \frac{6}{e ^{4}}) t - \frac{15}{e ^{4}} - 16 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( 20 + \frac{6}{e ^{4}} ) \pm ( 20 + \frac{6}{e ^{4}} ) ^{2} - 4 ( - 5 ) ( - \frac{15}{e ^{4}} - 16 )}{2 ( - 5 )}

(20 + \frac{6}{e ^{4}})^{2} - 4 (- 5) (- \frac{15}{e ^{4}} - 16) = 80 + \frac{36}{e ^{8}} - \frac{60}{e ^{4}}

t_{1, 2} = \frac{- ( 20 + \frac{6}{e ^{4}} ) \pm 80 + \frac{36}{e ^{8}} - \frac{60}{e ^{4}}}{2 ( - 5 )}

Separate the solutions

t_{1} = \frac{- ( 20 + \frac{6}{e ^{4}} ) + 80 + \frac{36}{e ^{8}} - \frac{60}{e ^{4}}}{2 ( - 5 )}, t_{2} = \frac{- ( 20 + \frac{6}{e ^{4}} ) - 80 + \frac{36}{e ^{8}} - \frac{60}{e ^{4}}}{2 ( - 5 )}

t = \frac{- ( 20 + \frac{6}{e ^{4}} ) + 80 + \frac{36}{e ^{8}} - \frac{60}{e ^{4}}}{2 ( - 5 )} : \frac{6 + 20 e ^{4} - e ^{4} 80 + \frac{36}{e ^{8}} - \frac{60}{e ^{4}}}{10 e ^{4}}

t = \frac{- ( 20 + \frac{6}{e ^{4}} ) - 80 + \frac{36}{e ^{8}} - \frac{60}{e ^{4}}}{2 ( - 5 )} : \frac{6 + 20 e ^{4} + e ^{4} 80 + \frac{36}{e ^{8}} - \frac{60}{e ^{4}}}{10 e ^{4}}

The solutions to the quadratic equation are:

t = \frac{6 + 20 e ^{4} - e ^{4} 80 + \frac{36}{e ^{8}} - \frac{60}{e ^{4}}}{10 e ^{4}}, t = \frac{6 + 20 e ^{4} + e ^{4} 80 + \frac{36}{e ^{8}} - \frac{60}{e ^{4}}}{10 e ^{4}}

x^{2} - \frac{3}{4} x - \frac{7}{64} = 0

A^{2} = A

x^{2} - 160 x + 10 = 0

35 x^{2} + 6 x - 8 = 0

- 2 y = 4 y^{2}

What are the solutions to the equation -5t^2+(20+6e^{-4})t-(16+15e^{-4})=0 ?
The solutions to the equation -5t^2+(20+6e^{-4})t-(16+15e^{-4})=0 are
Find the zeros of -5t^2+(20+6e^{-4})t-(16+15e^{-4})=0
The zeros of -5t^2+(20+6e^{-4})t-(16+15e^{-4})=0 are