What are the solutions to the equation 4-y^2=y-2 ?

The solutions to the equation 4-y^2=y-2 are y=-3,y=2

Find the zeros of 4-y^2=y-2

The zeros of 4-y^2=y-2 are y=-3,y=2

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4 - y^{2} = y - 2

y = - 3, y = 2

Solution steps

4 - y^{2} = y - 2

Move

2

to the left side

- y^{2} + 6 = y

Move

y

to the left side

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

Write in the standard form

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

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

Solve with the quadratic formula

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

(- 1)^{2} - 4 (- 1) \cdot 6 = 5

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

y = - 3, y = 2

8 a^{3} + 125 b^{3}

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6 x - x^{2}

5 k^{2} + 12 k + 7

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What are the solutions to the equation 4-y^2=y-2 ?
The solutions to the equation 4-y^2=y-2 are y=-3,y=2
Find the zeros of 4-y^2=y-2
The zeros of 4-y^2=y-2 are y=-3,y=2