What are the solutions to the equation z^2+z+25=0 ?

The solutions to the equation z^2+z+25=0 are z=-1/2+i(3sqrt(11))/2 ,z=-1/2-i(3sqrt(11))/2

Find the zeros of z^2+z+25=0

The zeros of z^2+z+25=0 are z=-1/2+i(3sqrt(11))/2 ,z=-1/2-i(3sqrt(11))/2

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z^{2} + z + 25 = 0

z = - \frac{1}{2} + i \frac{3 11}{2}, z = - \frac{1}{2} - i \frac{3 11}{2}

Solution steps

z^{2} + z + 25 = 0

Solve with the quadratic formula

z_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot 25}{2 \cdot 1}

Simplify

1^{2} - 4 \cdot 1 \cdot 25 : 311 i

z_{1, 2} = \frac{- 1 \pm 3 11 i}{2 \cdot 1}

Separate the solutions

z_{1} = \frac{- 1 + 3 11 i}{2 \cdot 1}, z_{2} = \frac{- 1 - 3 11 i}{2 \cdot 1}

z = \frac{- 1 + 3 11 i}{2 \cdot 1} : - \frac{1}{2} + i \frac{3 11}{2}

z = \frac{- 1 - 3 11 i}{2 \cdot 1} : - \frac{1}{2} - i \frac{3 11}{2}

The solutions to the quadratic equation are:

z = - \frac{1}{2} + i \frac{3 11}{2}, z = - \frac{1}{2} - i \frac{3 11}{2}

\frac{( \frac{3}{2} )}{\frac{1}{2}}

2 y^{2} - y - 15

4 x + 4 = 0

6 t^{2} - 12 t

4 x^{2} + 9 x^{2} = 36

What are the solutions to the equation z^2+z+25=0 ?
The solutions to the equation z^2+z+25=0 are z=-1/2+i(3sqrt(11))/2 ,z=-1/2-i(3sqrt(11))/2
Find the zeros of z^2+z+25=0
The zeros of z^2+z+25=0 are z=-1/2+i(3sqrt(11))/2 ,z=-1/2-i(3sqrt(11))/2