What are the solutions to the equation 0=32-2t-5t^2 ?

The solutions to the equation 0=32-2t-5t^2 are t=-(1+sqrt(161))/5 ,t=(sqrt(161)-1)/5

Find the zeros of 0=32-2t-5t^2

The zeros of 0=32-2t-5t^2 are t=-(1+sqrt(161))/5 ,t=(sqrt(161)-1)/5

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0 = 32 - 2 t - 5 t^{2}

t = - \frac{1 + 161}{5}, t = \frac{161 - 1}{5}

Decimal Notation

t = - 2.73771 \dots, t = 2.33771 \dots

Solution steps

0 = 32 - 2 t - 5 t^{2}

Switch sides

32 - 2 t - 5 t^{2} = 0

Write in the standard form

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

- 5 t^{2} - 2 t + 32 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 ( - 5 ) \cdot 32}{2 ( - 5 )}

(- 2)^{2} - 4 (- 5) \cdot 32 = 2161

t_{1, 2} = \frac{- ( - 2 ) \pm 2 161}{2 ( - 5 )}

Separate the solutions

t_{1} = \frac{- ( - 2 ) + 2 161}{2 ( - 5 )}, t_{2} = \frac{- ( - 2 ) - 2 161}{2 ( - 5 )}

t = \frac{- ( - 2 ) + 2 161}{2 ( - 5 )} : - \frac{1 + 161}{5}

t = \frac{- ( - 2 ) - 2 161}{2 ( - 5 )} : \frac{161 - 1}{5}

The solutions to the quadratic equation are:

t = - \frac{1 + 161}{5}, t = \frac{161 - 1}{5}

2 (x + 1)^{2} - 50 = 0

2 x^{2} + 4 = 5 x + 2

x^{2} - 12 x + 36 = 0

5 (5 x - 9)^{2} - 39 = 41

8 z^{2} + 13 z + 4 = 3 z^{2}

What are the solutions to the equation 0=32-2t-5t^2 ?
The solutions to the equation 0=32-2t-5t^2 are t=-(1+sqrt(161))/5 ,t=(sqrt(161)-1)/5
Find the zeros of 0=32-2t-5t^2
The zeros of 0=32-2t-5t^2 are t=-(1+sqrt(161))/5 ,t=(sqrt(161)-1)/5