What are the solutions to the equation 9k^2+24k+18=0 ?

The solutions to the equation 9k^2+24k+18=0 are k=-4/3+i(sqrt(2))/3 ,k=-4/3-i(sqrt(2))/3

Find the zeros of 9k^2+24k+18=0

The zeros of 9k^2+24k+18=0 are k=-4/3+i(sqrt(2))/3 ,k=-4/3-i(sqrt(2))/3

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9 k^{2} + 24 k + 18 = 0

k = - \frac{4}{3} + i \frac{2}{3}, k = - \frac{4}{3} - i \frac{2}{3}

Solution steps

9 k^{2} + 24 k + 18 = 0

Solve with the quadratic formula

k_{1, 2} = \frac{- 24 \pm 2 4 ^{2} - 4 \cdot 9 \cdot 18}{2 \cdot 9}

Simplify

2 4^{2} - 4 \cdot 9 \cdot 18 : 62 i

k_{1, 2} = \frac{- 24 \pm 6 2 i}{2 \cdot 9}

Separate the solutions

k_{1} = \frac{- 24 + 6 2 i}{2 \cdot 9}, k_{2} = \frac{- 24 - 6 2 i}{2 \cdot 9}

k = \frac{- 24 + 6 2 i}{2 \cdot 9} : - \frac{4}{3} + i \frac{2}{3}

k = \frac{- 24 - 6 2 i}{2 \cdot 9} : - \frac{4}{3} - i \frac{2}{3}

The solutions to the quadratic equation are:

k = - \frac{4}{3} + i \frac{2}{3}, k = - \frac{4}{3} - i \frac{2}{3}

\frac{( n + 4 ) !}{n !}

lo g_{3} (3^{- 2}) = x

f = (- 3)^{2} - 5

x^{2} + 3 x - 28 \leq 0

- 6 x - 21 - 5 x + 6 - 8 + 16 x - x + 3 = 0

What are the solutions to the equation 9k^2+24k+18=0 ?
The solutions to the equation 9k^2+24k+18=0 are k=-4/3+i(sqrt(2))/3 ,k=-4/3-i(sqrt(2))/3
Find the zeros of 9k^2+24k+18=0
The zeros of 9k^2+24k+18=0 are k=-4/3+i(sqrt(2))/3 ,k=-4/3-i(sqrt(2))/3