What are the solutions to the equation solvefor x,2x^2-kx+5=0 ?

The solutions to the equation solvefor x,2x^2-kx+5=0 are x=(k+sqrt(k^2-40))/4 ,x=(k-sqrt(k^2-40))/4

Find the zeros of solvefor x,2x^2-kx+5=0

The zeros of solvefor x,2x^2-kx+5=0 are x=(k+sqrt(k^2-40))/4 ,x=(k-sqrt(k^2-40))/4

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x, 2 x^{2} - k x + 5 = 0

x = \frac{k + k ^{2} - 40}{4}, x = \frac{k - k ^{2} - 40}{4}

Solution steps

2 x^{2} - k x + 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - k ) \pm ( - k ) ^{2} - 4 \cdot 2 \cdot 5}{2 \cdot 2}

Simplify

(- k)^{2} - 4 \cdot 2 \cdot 5 : k^{2} - 40

x_{1, 2} = \frac{- ( - k ) \pm k ^{2} - 40}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - k ) + k ^{2} - 40}{2 \cdot 2}, x_{2} = \frac{- ( - k ) - k ^{2} - 40}{2 \cdot 2}

x = \frac{- ( - k ) + k ^{2} - 40}{2 \cdot 2} : \frac{k + k ^{2} - 40}{4}

x = \frac{- ( - k ) - k ^{2} - 40}{2 \cdot 2} : \frac{k - k ^{2} - 40}{4}

The solutions to the quadratic equation are:

x = \frac{k + k ^{2} - 40}{4}, x = \frac{k - k ^{2} - 40}{4}

x^{2} - x - 73 = - x - 9

p, 0 = (500 - 10 p - q)^{2}

x^{2} - 3 x + (\frac{9}{4}) = 12

(2 x - 9)^{2} = 25

x, 7 x^{2} - 8 x y + 7 y^{2} - 12 = 0

What are the solutions to the equation solvefor x,2x^2-kx+5=0 ?
The solutions to the equation solvefor x,2x^2-kx+5=0 are x=(k+sqrt(k^2-40))/4 ,x=(k-sqrt(k^2-40))/4
Find the zeros of solvefor x,2x^2-kx+5=0
The zeros of solvefor x,2x^2-kx+5=0 are x=(k+sqrt(k^2-40))/4 ,x=(k-sqrt(k^2-40))/4