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

The solutions to the equation 4z^2+z+2=0 are z=-1/8+i(sqrt(31))/8 ,z=-1/8-i(sqrt(31))/8

Find the zeros of 4z^2+z+2=0

The zeros of 4z^2+z+2=0 are z=-1/8+i(sqrt(31))/8 ,z=-1/8-i(sqrt(31))/8

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4 z^{2} + z + 2 = 0

z = - \frac{1}{8} + i \frac{31}{8}, z = - \frac{1}{8} - i \frac{31}{8}

Solution steps

4 z^{2} + z + 2 = 0

Solve with the quadratic formula

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

Simplify

1^{2} - 4 \cdot 4 \cdot 2 : 31 i

z_{1, 2} = \frac{- 1 \pm 31 i}{2 \cdot 4}

Separate the solutions

z_{1} = \frac{- 1 + 31 i}{2 \cdot 4}, z_{2} = \frac{- 1 - 31 i}{2 \cdot 4}

z = \frac{- 1 + 31 i}{2 \cdot 4} : - \frac{1}{8} + i \frac{31}{8}

z = \frac{- 1 - 31 i}{2 \cdot 4} : - \frac{1}{8} - i \frac{31}{8}

The solutions to the quadratic equation are:

z = - \frac{1}{8} + i \frac{31}{8}, z = - \frac{1}{8} - i \frac{31}{8}

\frac{5}{9}

14 > 4 + 2 x

\frac{1}{5} x + 7 < 11

3 x - 5 \leq 4

6 - 6 ∣ - 7 + x ∣ > 3

What are the solutions to the equation 4z^2+z+2=0 ?
The solutions to the equation 4z^2+z+2=0 are z=-1/8+i(sqrt(31))/8 ,z=-1/8-i(sqrt(31))/8
Find the zeros of 4z^2+z+2=0
The zeros of 4z^2+z+2=0 are z=-1/8+i(sqrt(31))/8 ,z=-1/8-i(sqrt(31))/8