What are the solutions to the equation 4x^2+x+3=0 ?

The solutions to the equation 4x^2+x+3=0 are x=-1/8+i(sqrt(47))/8 ,x=-1/8-i(sqrt(47))/8

Find the zeros of 4x^2+x+3=0

The zeros of 4x^2+x+3=0 are x=-1/8+i(sqrt(47))/8 ,x=-1/8-i(sqrt(47))/8

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4 x^{2} + x + 3 = 0

x = - \frac{1}{8} + i \frac{47}{8}, x = - \frac{1}{8} - i \frac{47}{8}

Solution steps

4 x^{2} + x + 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 4 \cdot 3}{2 \cdot 4}

Simplify

1^{2} - 4 \cdot 4 \cdot 3 : 47 i

x_{1, 2} = \frac{- 1 \pm 47 i}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 1 + 47 i}{2 \cdot 4}, x_{2} = \frac{- 1 - 47 i}{2 \cdot 4}

x = \frac{- 1 + 47 i}{2 \cdot 4} : - \frac{1}{8} + i \frac{47}{8}

x = \frac{- 1 - 47 i}{2 \cdot 4} : - \frac{1}{8} - i \frac{47}{8}

The solutions to the quadratic equation are:

x = - \frac{1}{8} + i \frac{47}{8}, x = - \frac{1}{8} - i \frac{47}{8}

- 10 < - 2 x \leq 10

8 x^{4} + 8 x^{3} + 10 x^{2} + 8 x + 2

2^{2} = 8

2 y^{3} - 128

\frac{x}{8} + 1 = 6

What are the solutions to the equation 4x^2+x+3=0 ?
The solutions to the equation 4x^2+x+3=0 are x=-1/8+i(sqrt(47))/8 ,x=-1/8-i(sqrt(47))/8
Find the zeros of 4x^2+x+3=0
The zeros of 4x^2+x+3=0 are x=-1/8+i(sqrt(47))/8 ,x=-1/8-i(sqrt(47))/8