What are the solutions to the equation 2k(k+7)=-49 ?

The solutions to the equation 2k(k+7)=-49 are k=-7/2+i 7/2 ,k=-7/2-i 7/2

Find the zeros of 2k(k+7)=-49

The zeros of 2k(k+7)=-49 are k=-7/2+i 7/2 ,k=-7/2-i 7/2

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2 k (k + 7) = - 49

k = - \frac{7}{2} + i \frac{7}{2}, k = - \frac{7}{2} - i \frac{7}{2}

Solution steps

2 k (k + 7) = - 49

Expand

2 k (k + 7) : 2 k^{2} + 14 k

2 k^{2} + 14 k = - 49

Move

49

to the left side

2 k^{2} + 14 k + 49 = 0

Solve with the quadratic formula

k_{1, 2} = \frac{- 14 \pm 1 4 ^{2} - 4 \cdot 2 \cdot 49}{2 \cdot 2}

Simplify

1 4^{2} - 4 \cdot 2 \cdot 49 : 14 i

k_{1, 2} = \frac{- 14 \pm 14 i}{2 \cdot 2}

Separate the solutions

k_{1} = \frac{- 14 + 14 i}{2 \cdot 2}, k_{2} = \frac{- 14 - 14 i}{2 \cdot 2}

k = \frac{- 14 + 14 i}{2 \cdot 2} : - \frac{7}{2} + i \frac{7}{2}

k = \frac{- 14 - 14 i}{2 \cdot 2} : - \frac{7}{2} - i \frac{7}{2}

The solutions to the quadratic equation are:

k = - \frac{7}{2} + i \frac{7}{2}, k = - \frac{7}{2} - i \frac{7}{2}

7 n^{2} - 3 n - 3 = 0

6 r^{2} - 64 + 5 r = 2 r^{2} + 5 r

z^{2} - z + 2 = 0

2 x^{2} + 8 x = 54

0 = 3 x^{2} + 16 x + 5

What are the solutions to the equation 2k(k+7)=-49 ?
The solutions to the equation 2k(k+7)=-49 are k=-7/2+i 7/2 ,k=-7/2-i 7/2
Find the zeros of 2k(k+7)=-49
The zeros of 2k(k+7)=-49 are k=-7/2+i 7/2 ,k=-7/2-i 7/2