What are the solutions to the equation solvefor x,x^2-kx-2=0 ?

The solutions to the equation solvefor x,x^2-kx-2=0 are x=(k+sqrt(k^2+8))/2 ,x=(k-sqrt(k^2+8))/2

Find the zeros of solvefor x,x^2-kx-2=0

The zeros of solvefor x,x^2-kx-2=0 are x=(k+sqrt(k^2+8))/2 ,x=(k-sqrt(k^2+8))/2

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x, x^{2} - k x - 2 = 0

x = \frac{k + k ^{2} + 8}{2}, x = \frac{k - k ^{2} + 8}{2}

Solution steps

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

Solve with the quadratic formula

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

Simplify

(- k)^{2} - 4 \cdot 1 \cdot (- 2) : k^{2} + 8

x_{1, 2} = \frac{- ( - k ) \pm k ^{2} + 8}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - k ) + k ^{2} + 8}{2 \cdot 1}, x_{2} = \frac{- ( - k ) - k ^{2} + 8}{2 \cdot 1}

x = \frac{- ( - k ) + k ^{2} + 8}{2 \cdot 1} : \frac{k + k ^{2} + 8}{2}

x = \frac{- ( - k ) - k ^{2} + 8}{2 \cdot 1} : \frac{k - k ^{2} + 8}{2}

The solutions to the quadratic equation are:

x = \frac{k + k ^{2} + 8}{2}, x = \frac{k - k ^{2} + 8}{2}

81 - x^{2} = 2 x^{2}

2 (x + 5)^{2} - 6 = 0

x^{2} - (5 - 3 x) = x + 10

x^{2} - 9 x = x + 24

2 x^{2} + 5 x = 0

What are the solutions to the equation solvefor x,x^2-kx-2=0 ?
The solutions to the equation solvefor x,x^2-kx-2=0 are x=(k+sqrt(k^2+8))/2 ,x=(k-sqrt(k^2+8))/2
Find the zeros of solvefor x,x^2-kx-2=0
The zeros of solvefor x,x^2-kx-2=0 are x=(k+sqrt(k^2+8))/2 ,x=(k-sqrt(k^2+8))/2