What are the solutions to the equation solvefor x,x^2+Kx+9=0 ?

The solutions to the equation solvefor x,x^2+Kx+9=0 are x=(-K+sqrt(K^2-36))/2 ,x=(-K-sqrt(K^2-36))/2

Find the zeros of solvefor x,x^2+Kx+9=0

The zeros of solvefor x,x^2+Kx+9=0 are x=(-K+sqrt(K^2-36))/2 ,x=(-K-sqrt(K^2-36))/2

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solve for

x, x^{2} + K x + 9 = 0

x = \frac{- K + K ^{2} - 36}{2}, x = \frac{- K - K ^{2} - 36}{2}

Solution steps

x^{2} + K x + 9 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- K \pm K ^{2} - 4 \cdot 1 \cdot 9}{2 \cdot 1}

Simplify

K^{2} - 4 \cdot 1 \cdot 9 : K^{2} - 36

x_{1, 2} = \frac{- K \pm K ^{2} - 36}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- K + K ^{2} - 36}{2 \cdot 1}, x_{2} = \frac{- K - K ^{2} - 36}{2 \cdot 1}

x = \frac{- K + K ^{2} - 36}{2 \cdot 1} : \frac{- K + K ^{2} - 36}{2}

x = \frac{- K - K ^{2} - 36}{2 \cdot 1} : \frac{- K - K ^{2} - 36}{2}

The solutions to the quadratic equation are:

x = \frac{- K + K ^{2} - 36}{2}, x = \frac{- K - K ^{2} - 36}{2}

x, 9 x^{2} - 4 y^{2} - z^{2} + 16 y - 17 = 0

16 x^{2} - 23 x = 17 x - 25

9 y^{2} - y = 0

x, 3 x y - 5 x^{2} - y^{2} = 3

x^{2} - 4 x - 26 = - 2

What are the solutions to the equation solvefor x,x^2+Kx+9=0 ?
The solutions to the equation solvefor x,x^2+Kx+9=0 are x=(-K+sqrt(K^2-36))/2 ,x=(-K-sqrt(K^2-36))/2
Find the zeros of solvefor x,x^2+Kx+9=0
The zeros of solvefor x,x^2+Kx+9=0 are x=(-K+sqrt(K^2-36))/2 ,x=(-K-sqrt(K^2-36))/2