What are the solutions to the equation 5t^2+7t-1=0 ?

The solutions to the equation 5t^2+7t-1=0 are t=(-7+sqrt(69))/(10),t=(-7-sqrt(69))/(10)

Find the zeros of 5t^2+7t-1=0

The zeros of 5t^2+7t-1=0 are t=(-7+sqrt(69))/(10),t=(-7-sqrt(69))/(10)

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

t = \frac{- 7 + 69}{10}, t = \frac{- 7 - 69}{10}

Decimal Notation

t = 0.13066 \dots, t = - 1.53066 \dots

Solution steps

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

Solve with the quadratic formula

t_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 5 ( - 1 )}{2 \cdot 5}

7^{2} - 4 \cdot 5 (- 1) = 69

t_{1, 2} = \frac{- 7 \pm 69}{2 \cdot 5}

Separate the solutions

t_{1} = \frac{- 7 + 69}{2 \cdot 5}, t_{2} = \frac{- 7 - 69}{2 \cdot 5}

t = \frac{- 7 + 69}{2 \cdot 5} : \frac{- 7 + 69}{10}

t = \frac{- 7 - 69}{2 \cdot 5} : \frac{- 7 - 69}{10}

The solutions to the quadratic equation are:

t = \frac{- 7 + 69}{10}, t = \frac{- 7 - 69}{10}

y, y^{2} - x^{2} + 1 = 50

12 + y^{2} - 6 y = 0

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

5 x^{2} - 24 x + 16 = 0

3 v^{2} = v + 10

What are the solutions to the equation 5t^2+7t-1=0 ?
The solutions to the equation 5t^2+7t-1=0 are t=(-7+sqrt(69))/(10),t=(-7-sqrt(69))/(10)
Find the zeros of 5t^2+7t-1=0
The zeros of 5t^2+7t-1=0 are t=(-7+sqrt(69))/(10),t=(-7-sqrt(69))/(10)