What are the solutions to the equation k^2-2k-8=0 ?

The solutions to the equation k^2-2k-8=0 are k=4,k=-2

Find the zeros of k^2-2k-8=0

The zeros of k^2-2k-8=0 are k=4,k=-2

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k^{2} - 2 k - 8 = 0

k = 4, k = - 2

Solution steps

k^{2} - 2 k - 8 = 0

Solve with the quadratic formula

k_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot ( - 8 )}{2 \cdot 1}

(- 2)^{2} - 4 \cdot 1 \cdot (- 8) = 6

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

Separate the solutions

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

k = \frac{- ( - 2 ) + 6}{2 \cdot 1} : 4

k = \frac{- ( - 2 ) - 6}{2 \cdot 1} : - 2

The solutions to the quadratic equation are:

k = 4, k = - 2

\frac{3 x ^{5} y}{12 x ^{3} y ^{2}}

5 x^{2} - 35 = 0

2 (0) - 5 y = 10

x^{2} - 34 x - 1827

6 c^{2} + 3 = 51

What are the solutions to the equation k^2-2k-8=0 ?
The solutions to the equation k^2-2k-8=0 are k=4,k=-2
Find the zeros of k^2-2k-8=0
The zeros of k^2-2k-8=0 are k=4,k=-2