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

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

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

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

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

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

Decimal Notation

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

Solution steps

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

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

The solutions to the quadratic equation are:

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

5 x^{2} + 12 x + 3 = 0

3 x^{2} + 5 x - 182 = 0

(x - 3) (x + 7) + (x - 3) (x - 4) = 0

(2 x + 1)^{2} = (5 x - 1) (5 x + 1)

r^{2} = (3)^{2} + (- 1)^{2}

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