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

The solutions to the equation k^2-k-2=0 are k=2,k=-1

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

The zeros of k^2-k-2=0 are k=2,k=-1

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k^{2} - k - 2 = 0

k = 2, k = - 1

Solution steps

k^{2} - k - 2 = 0

Solve with the quadratic formula

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

(- 1)^{2} - 4 \cdot 1 \cdot (- 2) = 3

k_{1, 2} = \frac{- ( - 1 ) \pm 3}{2 \cdot 1}

Separate the solutions

k_{1} = \frac{- ( - 1 ) + 3}{2 \cdot 1}, k_{2} = \frac{- ( - 1 ) - 3}{2 \cdot 1}

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

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

The solutions to the quadratic equation are:

k = 2, k = - 1

x^{18} - y^{15}

a = π (8.5)^{2}

- 12 + 7 - 9 \cdot 3

\frac{2 ( x + 1 )}{4} = 3 x - 2

\frac{2}{3} (2 + 2)^{\frac{2}{3}}

What are the solutions to the equation k^2-k-2=0 ?
The solutions to the equation k^2-k-2=0 are k=2,k=-1
Find the zeros of k^2-k-2=0
The zeros of k^2-k-2=0 are k=2,k=-1