What are the solutions to the equation 4-4y-y^2=0 ?

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

Find the zeros of 4-4y-y^2=0

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

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4 - 4 y - y^{2} = 0

y = - 2 (1 + 2), y = 2 (2 - 1)

Decimal Notation

y = - 4.82842 \dots, y = 0.82842 \dots

Solution steps

4 - 4 y - y^{2} = 0

Write in the standard form

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

- y^{2} - 4 y + 4 = 0

Solve with the quadratic formula

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

(- 4)^{2} - 4 (- 1) \cdot 4 = 42

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

Separate the solutions

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

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

y = \frac{- ( - 4 ) - 4 2}{2 ( - 1 )} : 2 (2 - 1)

The solutions to the quadratic equation are:

y = - 2 (1 + 2), y = 2 (2 - 1)

7 x^{2} = - 4 x

2 λ^{2} - 1 = 0

x, z = 1 + (x - 1)^{2} + 4 y^{2}

m^{2} - 5 m - 14 = 0

- 10000 x^{2} + 4000 x + 9000 = 0

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