What are the solutions to the equation 5t^2-2t-1=0 ?

The solutions to the equation 5t^2-2t-1=0 are t=(1+sqrt(6))/5 ,t=(1-sqrt(6))/5

Find the zeros of 5t^2-2t-1=0

The zeros of 5t^2-2t-1=0 are t=(1+sqrt(6))/5 ,t=(1-sqrt(6))/5

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5 t^{2} - 2 t - 1 = 0

t = \frac{1 + 6}{5}, t = \frac{1 - 6}{5}

Decimal Notation

t = 0.68989 \dots, t = - 0.28989 \dots

Solution steps

5 t^{2} - 2 t - 1 = 0

Solve with the quadratic formula

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

(- 2)^{2} - 4 \cdot 5 (- 1) = 26

t_{1, 2} = \frac{- ( - 2 ) \pm 2 6}{2 \cdot 5}

Separate the solutions

t_{1} = \frac{- ( - 2 ) + 2 6}{2 \cdot 5}, t_{2} = \frac{- ( - 2 ) - 2 6}{2 \cdot 5}

t = \frac{- ( - 2 ) + 2 6}{2 \cdot 5} : \frac{1 + 6}{5}

t = \frac{- ( - 2 ) - 2 6}{2 \cdot 5} : \frac{1 - 6}{5}

The solutions to the quadratic equation are:

t = \frac{1 + 6}{5}, t = \frac{1 - 6}{5}

\frac{9 x ^{2} + 3 x}{4 + 2} = 0

\frac{x ^{2}}{2 + 2} = 20

\frac{0.44 x ( x + x \cdot 5 - \frac{23}{55} ) + 3}{14 \cdot 2.24} = 1

48 x^{2} - 13 x - 1 = 0

(k^{2} + 3) \cdot x^{2} + (k^{3} + k + 1) \cdot x + 2 k + 2 = 0

What are the solutions to the equation 5t^2-2t-1=0 ?
The solutions to the equation 5t^2-2t-1=0 are t=(1+sqrt(6))/5 ,t=(1-sqrt(6))/5
Find the zeros of 5t^2-2t-1=0
The zeros of 5t^2-2t-1=0 are t=(1+sqrt(6))/5 ,t=(1-sqrt(6))/5