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

The solutions to the equation solvefor y,y^2-y-x^2=0 are y=(1+sqrt(1+4x^2))/2 ,y=(1-sqrt(1+4x^2))/2

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

The zeros of solvefor y,y^2-y-x^2=0 are y=(1+sqrt(1+4x^2))/2 ,y=(1-sqrt(1+4x^2))/2

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solve for

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

y = \frac{1 + 1 + 4 x ^{2}}{2}, y = \frac{1 - 1 + 4 x ^{2}}{2}

Solution steps

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

Solve with the quadratic formula

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

Simplify

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

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

Separate the solutions

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

y = \frac{- ( - 1 ) + 1 + 4 x ^{2}}{2 \cdot 1} : \frac{1 + 1 + 4 x ^{2}}{2}

y = \frac{- ( - 1 ) - 1 + 4 x ^{2}}{2 \cdot 1} : \frac{1 - 1 + 4 x ^{2}}{2}

The solutions to the quadratic equation are:

y = \frac{1 + 1 + 4 x ^{2}}{2}, y = \frac{1 - 1 + 4 x ^{2}}{2}

2 (x - 1)^{2} - (x + 4) (x - 4) = 19 - x

2 x^{2} - 3 x - 10

2 x^{2} - 12 x + 16 = 10

x^{2} + 13 = 6 x

(6 x - 1)^{2} + 4 = 0

What are the solutions to the equation solvefor y,y^2-y-x^2=0 ?
The solutions to the equation solvefor y,y^2-y-x^2=0 are y=(1+sqrt(1+4x^2))/2 ,y=(1-sqrt(1+4x^2))/2
Find the zeros of solvefor y,y^2-y-x^2=0
The zeros of solvefor y,y^2-y-x^2=0 are y=(1+sqrt(1+4x^2))/2 ,y=(1-sqrt(1+4x^2))/2