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

The solutions to the equation 4x^2+x-4=0 are x=(-1+sqrt(65))/8 ,x=(-1-sqrt(65))/8

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

The zeros of 4x^2+x-4=0 are x=(-1+sqrt(65))/8 ,x=(-1-sqrt(65))/8

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

x = \frac{- 1 + 65}{8}, x = \frac{- 1 - 65}{8}

Solution steps

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

Solve with the quadratic formula

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

1^{2} - 4 \cdot 4 (- 4) = 65

x_{1, 2} = \frac{- 1 \pm 65}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 1 + 65}{2 \cdot 4}, x_{2} = \frac{- 1 - 65}{2 \cdot 4}

x = \frac{- 1 + 65}{2 \cdot 4} : \frac{- 1 + 65}{8}

x = \frac{- 1 - 65}{2 \cdot 4} : \frac{- 1 - 65}{8}

The solutions to the quadratic equation are:

x = \frac{- 1 + 65}{8}, x = \frac{- 1 - 65}{8}

1.2 \cdot 5

\frac{9}{2} - 2

6 x + 7 < 12 x - 10

8 x + 16 = 6 x

x^{2} + 1 > 0

What are the solutions to the equation 4x^2+x-4=0 ?
The solutions to the equation 4x^2+x-4=0 are x=(-1+sqrt(65))/8 ,x=(-1-sqrt(65))/8
Find the zeros of 4x^2+x-4=0
The zeros of 4x^2+x-4=0 are x=(-1+sqrt(65))/8 ,x=(-1-sqrt(65))/8