What are the solutions to the equation α^2+3α+9=0 ?

The solutions to the equation α^2+3α+9=0 are α=-3/2+i(3sqrt(3))/2 ,α=-3/2-i(3sqrt(3))/2

Find the zeros of α^2+3α+9=0

The zeros of α^2+3α+9=0 are α=-3/2+i(3sqrt(3))/2 ,α=-3/2-i(3sqrt(3))/2

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α^{2} + 3 α + 9 = 0

α = - \frac{3}{2} + i \frac{3 3}{2}, α = - \frac{3}{2} - i \frac{3 3}{2}

Solution steps

α^{2} + 3 α + 9 = 0

Solve with the quadratic formula

α_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot 9}{2 \cdot 1}

Simplify

3^{2} - 4 \cdot 1 \cdot 9 : 33 i

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

α = - \frac{3}{2} + i \frac{3 3}{2}, α = - \frac{3}{2} - i \frac{3 3}{2}

x^{2} - 5 x + 10 = 4

4 b^{2} - 19 b + 20 = 0

6 x (x - 1) = 4 x + 5

(\frac{1}{8} - x)^{2} + 42 = (x + \frac{1}{8})^{2} - \frac{x}{2} + x (x - 1)

2 q^{2} - 3 q - 15 = - 4 q

What are the solutions to the equation α^2+3α+9=0 ?
The solutions to the equation α^2+3α+9=0 are α=-3/2+i(3sqrt(3))/2 ,α=-3/2-i(3sqrt(3))/2
Find the zeros of α^2+3α+9=0
The zeros of α^2+3α+9=0 are α=-3/2+i(3sqrt(3))/2 ,α=-3/2-i(3sqrt(3))/2