What are the solutions to the equation 2k(k+3)=-9 ?

The solutions to the equation 2k(k+3)=-9 are k=-3/2+i 3/2 ,k=-3/2-i 3/2

Find the zeros of 2k(k+3)=-9

The zeros of 2k(k+3)=-9 are k=-3/2+i 3/2 ,k=-3/2-i 3/2

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2 k (k + 3) = - 9

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

Solution steps

2 k (k + 3) = - 9

Expand

2 k (k + 3) : 2 k^{2} + 6 k

2 k^{2} + 6 k = - 9

Move

9

to the left side

2 k^{2} + 6 k + 9 = 0

Solve with the quadratic formula

k_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 2 \cdot 9}{2 \cdot 2}

Simplify

6^{2} - 4 \cdot 2 \cdot 9 : 6 i

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

2 x^{2} + 8 x - 1 = 0

2 x^{2} - 2 x - 10 = 0

4 x^{2} + 5 = 12

\frac{1}{8} x^{2} + \frac{5}{4} x + \frac{1}{3} = 0

4 x^{2} + 5 = - 7

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