What are the solutions to the equation -k^2-k+6=0 ?

The solutions to the equation -k^2-k+6=0 are k=-3,k=2

Find the zeros of -k^2-k+6=0

The zeros of -k^2-k+6=0 are k=-3,k=2

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

k = - 3, k = 2

Solution steps

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

Solve with the quadratic formula

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

(- 1)^{2} - 4 (- 1) \cdot 6 = 5

k_{1, 2} = \frac{- ( - 1 ) \pm 5}{2 ( - 1 )}

Separate the solutions

k_{1} = \frac{- ( - 1 ) + 5}{2 ( - 1 )}, k_{2} = \frac{- ( - 1 ) - 5}{2 ( - 1 )}

k = \frac{- ( - 1 ) + 5}{2 ( - 1 )} : - 3

k = \frac{- ( - 1 ) - 5}{2 ( - 1 )} : 2

The solutions to the quadratic equation are:

k = - 3, k = 2

2 0^{2} + 6^{2} = c^{2}

5 a^{2} + 2 a - 33 = 0

a^{2} + 3 a + 1 = 0

b^{2} + 8 b + 7 = 0

(1) = x^{2}

What are the solutions to the equation -k^2-k+6=0 ?
The solutions to the equation -k^2-k+6=0 are k=-3,k=2
Find the zeros of -k^2-k+6=0
The zeros of -k^2-k+6=0 are k=-3,k=2