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

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

Find the zeros of y^2=2+y

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

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y^{2} = 2 + y

y = 2, y = - 1

Solution steps

y^{2} = 2 + y

Move

y

to the left side

y^{2} - y = 2

Move

2

to the left side

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} : 2

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

The solutions to the quadratic equation are:

y = 2, y = - 1

8192 - 27

2 x + 5 + 2 x + 3 x = 2 x + 9

3 x^{2} + 23 x + 14

2 (6 + 4 n) \geq 12 - 8 n

- \frac{1}{3} x^{2} - 1 = 29

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