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

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

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

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

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

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

Decimal Notation

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

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

3 x^{2} + 6 x - 72 = 0

15 x^{2} - 4 x y - 4 y^{2}

3 x^{2} + x = 2

(\frac{3}{5})^{- 1}

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