What are the solutions to the equation -4+x+7x^2=0 ?

The solutions to the equation -4+x+7x^2=0 are x=(-1+sqrt(113))/(14),x=(-1-sqrt(113))/(14)

Find the zeros of -4+x+7x^2=0

The zeros of -4+x+7x^2=0 are x=(-1+sqrt(113))/(14),x=(-1-sqrt(113))/(14)

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- 4 + x + 7 x^{2} = 0

x = \frac{- 1 + 113}{14}, x = \frac{- 1 - 113}{14}

Solution steps

- 4 + x + 7 x^{2} = 0

Write in the standard form

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

7 x^{2} + x - 4 = 0

Solve with the quadratic formula

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

1^{2} - 4 \cdot 7 (- 4) = 113

x_{1, 2} = \frac{- 1 \pm 113}{2 \cdot 7}

Separate the solutions

x_{1} = \frac{- 1 + 113}{2 \cdot 7}, x_{2} = \frac{- 1 - 113}{2 \cdot 7}

x = \frac{- 1 + 113}{2 \cdot 7} : \frac{- 1 + 113}{14}

x = \frac{- 1 - 113}{2 \cdot 7} : \frac{- 1 - 113}{14}

The solutions to the quadratic equation are:

x = \frac{- 1 + 113}{14}, x = \frac{- 1 - 113}{14}

\frac{1}{100} + \frac{1}{200} + \frac{1}{300}

3 x + 12 - 6 x \leq - 9

(\frac{1}{8})^{3}

12 a^{2} + 16 ab - 3 b^{2}

2 t^{2} - 12 t = 12

What are the solutions to the equation -4+x+7x^2=0 ?
The solutions to the equation -4+x+7x^2=0 are x=(-1+sqrt(113))/(14),x=(-1-sqrt(113))/(14)
Find the zeros of -4+x+7x^2=0
The zeros of -4+x+7x^2=0 are x=(-1+sqrt(113))/(14),x=(-1-sqrt(113))/(14)