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

The solutions to the equation t^2+5t-1=0 are t=(-5+sqrt(29))/2 ,t=(-5-sqrt(29))/2

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

The zeros of t^2+5t-1=0 are t=(-5+sqrt(29))/2 ,t=(-5-sqrt(29))/2

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

t = \frac{- 5 + 29}{2}, t = \frac{- 5 - 29}{2}

Decimal Notation

t = 0.19258 \dots, t = - 5.19258 \dots

Solution steps

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

Solve with the quadratic formula

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

5^{2} - 4 \cdot 1 \cdot (- 1) = 29

t_{1, 2} = \frac{- 5 \pm 29}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- 5 + 29}{2 \cdot 1}, t_{2} = \frac{- 5 - 29}{2 \cdot 1}

t = \frac{- 5 + 29}{2 \cdot 1} : \frac{- 5 + 29}{2}

t = \frac{- 5 - 29}{2 \cdot 1} : \frac{- 5 - 29}{2}

The solutions to the quadratic equation are:

t = \frac{- 5 + 29}{2}, t = \frac{- 5 - 29}{2}

4 x^{2} + 16 x < 0

3 (4 + f) = 3 f + 12

4 \cdot \frac{3}{8}

\frac{4}{5} - \frac{2 p}{5} = \frac{6}{5}

(\frac{4}{5})^{1}

What are the solutions to the equation t^2+5t-1=0 ?
The solutions to the equation t^2+5t-1=0 are t=(-5+sqrt(29))/2 ,t=(-5-sqrt(29))/2
Find the zeros of t^2+5t-1=0
The zeros of t^2+5t-1=0 are t=(-5+sqrt(29))/2 ,t=(-5-sqrt(29))/2