What are the solutions to the equation solvefor y,y^2-xy=6 ?

The solutions to the equation solvefor y,y^2-xy=6 are y=(x+sqrt(x^2+24))/2 ,y=(x-sqrt(x^2+24))/2

Find the zeros of solvefor y,y^2-xy=6

The zeros of solvefor y,y^2-xy=6 are y=(x+sqrt(x^2+24))/2 ,y=(x-sqrt(x^2+24))/2

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solve for

y, y^{2} - x y = 6

y = \frac{x + x ^{2} + 24}{2}, y = \frac{x - x ^{2} + 24}{2}

Solution steps

y^{2} - x y = 6

Move

6

to the left side

y^{2} - x y - 6 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - x ) \pm ( - x ) ^{2} - 4 \cdot 1 \cdot ( - 6 )}{2 \cdot 1}

Simplify

(- x)^{2} - 4 \cdot 1 \cdot (- 6) : x^{2} + 24

y_{1, 2} = \frac{- ( - x ) \pm x ^{2} + 24}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- ( - x ) + x ^{2} + 24}{2 \cdot 1}, y_{2} = \frac{- ( - x ) - x ^{2} + 24}{2 \cdot 1}

y = \frac{- ( - x ) + x ^{2} + 24}{2 \cdot 1} : \frac{x + x ^{2} + 24}{2}

y = \frac{- ( - x ) - x ^{2} + 24}{2 \cdot 1} : \frac{x - x ^{2} + 24}{2}

The solutions to the quadratic equation are:

y = \frac{x + x ^{2} + 24}{2}, y = \frac{x - x ^{2} + 24}{2}

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

- 4 x^{2} + 16 x + 9 = 0

4 x^{2} - 65 x + 16 = 0

x (x + 3) = 70

z^{2} = (1 - i)^{\frac{1}{3}}

What are the solutions to the equation solvefor y,y^2-xy=6 ?
The solutions to the equation solvefor y,y^2-xy=6 are y=(x+sqrt(x^2+24))/2 ,y=(x-sqrt(x^2+24))/2
Find the zeros of solvefor y,y^2-xy=6
The zeros of solvefor y,y^2-xy=6 are y=(x+sqrt(x^2+24))/2 ,y=(x-sqrt(x^2+24))/2