What are the solutions to the equation y-y^2=y^2-3 ?

The solutions to the equation y-y^2=y^2-3 are y=-1,y= 3/2

Find the zeros of y-y^2=y^2-3

The zeros of y-y^2=y^2-3 are y=-1,y= 3/2

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y - y^{2} = y^{2} - 3

y = - 1, y = \frac{3}{2}

Decimal Notation

y = - 1, y = 1.5

Solution steps

y - y^{2} = y^{2} - 3

Move

3

to the left side

y - y^{2} + 3 = y^{2}

Move

y^{2}

to the left side

- 2 y^{2} + y + 3 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 2 ) \cdot 3}{2 ( - 2 )}

1^{2} - 4 (- 2) \cdot 3 = 5

y_{1, 2} = \frac{- 1 \pm 5}{2 ( - 2 )}

Separate the solutions

y_{1} = \frac{- 1 + 5}{2 ( - 2 )}, y_{2} = \frac{- 1 - 5}{2 ( - 2 )}

y = \frac{- 1 + 5}{2 ( - 2 )} : - 1

y = \frac{- 1 - 5}{2 ( - 2 )} : \frac{3}{2}

The solutions to the quadratic equation are:

y = - 1, y = \frac{3}{2}

y^{2} = - 16 y

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

3 (2 x + 1)^{2} + 2 = 7

w^{2} + 50 = - 15 w

4 x^{2} + 7 x - 7 = 0

What are the solutions to the equation y-y^2=y^2-3 ?
The solutions to the equation y-y^2=y^2-3 are y=-1,y= 3/2
Find the zeros of y-y^2=y^2-3
The zeros of y-y^2=y^2-3 are y=-1,y= 3/2