What are the solutions to the equation t^2+7t-1=0 ?

The solutions to the equation t^2+7t-1=0 are t=(-7+sqrt(53))/2 ,t=(-7-sqrt(53))/2

Find the zeros of t^2+7t-1=0

The zeros of t^2+7t-1=0 are t=(-7+sqrt(53))/2 ,t=(-7-sqrt(53))/2

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t^{2} + 7 t - 1 = 0

t = \frac{- 7 + 53}{2}, t = \frac{- 7 - 53}{2}

Decimal Notation

t = 0.14005 \dots, t = - 7.14005 \dots

Solution steps

t^{2} + 7 t - 1 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 1 \cdot ( - 1 )}{2 \cdot 1}

7^{2} - 4 \cdot 1 \cdot (- 1) = 53

t_{1, 2} = \frac{- 7 \pm 53}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- 7 + 53}{2 \cdot 1}, t_{2} = \frac{- 7 - 53}{2 \cdot 1}

t = \frac{- 7 + 53}{2 \cdot 1} : \frac{- 7 + 53}{2}

t = \frac{- 7 - 53}{2 \cdot 1} : \frac{- 7 - 53}{2}

The solutions to the quadratic equation are:

t = \frac{- 7 + 53}{2}, t = \frac{- 7 - 53}{2}

x^{2} - 27 x + 50 = 0

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625 t^{2} + 100 t^{1} + 4

a^{4} + 4 a^{2} m + 4 m^{2}

What are the solutions to the equation t^2+7t-1=0 ?
The solutions to the equation t^2+7t-1=0 are t=(-7+sqrt(53))/2 ,t=(-7-sqrt(53))/2
Find the zeros of t^2+7t-1=0
The zeros of t^2+7t-1=0 are t=(-7+sqrt(53))/2 ,t=(-7-sqrt(53))/2