What are the solutions to the equation y^2-2=-y ?

The solutions to the equation y^2-2=-y are y=1,y=-2

Find the zeros of y^2-2=-y

The zeros of y^2-2=-y are y=1,y=-2

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y^{2} - 2 = - y

y = 1, y = - 2

Solution steps

y^{2} - 2 = - y

Move

y

to the left side

y^{2} - 2 + y = 0

Write in the standard form

a x^{2} + b x + c = 0

y^{2} + y - 2 = 0

Solve with the quadratic formula

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

1^{2} - 4 \cdot 1 \cdot (- 2) = 3

y_{1, 2} = \frac{- 1 \pm 3}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- 1 + 3}{2 \cdot 1}, y_{2} = \frac{- 1 - 3}{2 \cdot 1}

y = \frac{- 1 + 3}{2 \cdot 1} : 1

y = \frac{- 1 - 3}{2 \cdot 1} : - 2

The solutions to the quadratic equation are:

y = 1, y = - 2

6 a^{2} + 8 a - 9 = 0

x^{2} + 5 x - 32 = 0

x^{2} - 9 x + 13 = 2 x + 3

5 x^{2} - 21 x - 5 = - 7 x - 2

5 x^{2} + 9 x - 6 = 10 x

What are the solutions to the equation y^2-2=-y ?
The solutions to the equation y^2-2=-y are y=1,y=-2
Find the zeros of y^2-2=-y
The zeros of y^2-2=-y are y=1,y=-2