What are the solutions to the equation 4+y^2=8+4y ?

The solutions to the equation 4+y^2=8+4y are y=2(1+sqrt(2)),y=2(1-sqrt(2))

Find the zeros of 4+y^2=8+4y

The zeros of 4+y^2=8+4y are y=2(1+sqrt(2)),y=2(1-sqrt(2))

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4 + y^{2} = 8 + 4 y

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

Decimal Notation

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

Solution steps

4 + y^{2} = 8 + 4 y

Move

4 y

to the left side

4 + y^{2} - 4 y = 8

Move

8

to the left side

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)

x^{2} + 23 x - 24 = 0

400 = 6 Q^{2} - 150 Q + 1000

(x - 10) (x) = 600

3 x^{2} + 9 x + 1

10 x^{2} - 40 x - 960 = 0

What are the solutions to the equation 4+y^2=8+4y ?
The solutions to the equation 4+y^2=8+4y are y=2(1+sqrt(2)),y=2(1-sqrt(2))
Find the zeros of 4+y^2=8+4y
The zeros of 4+y^2=8+4y are y=2(1+sqrt(2)),y=2(1-sqrt(2))