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

The solutions to the equation y^2-2y-2=0 are y=1+sqrt(3),y=1-sqrt(3)

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

The zeros of y^2-2y-2=0 are y=1+sqrt(3),y=1-sqrt(3)

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

y = 1 + 3, y = 1 - 3

Decimal Notation

y = 2.73205 \dots, y = - 0.73205 \dots

Solution steps

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

Solve with the quadratic formula

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

(- 2)^{2} - 4 \cdot 1 \cdot (- 2) = 23

y_{1, 2} = \frac{- ( - 2 ) \pm 2 3}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- ( - 2 ) + 2 3}{2 \cdot 1}, y_{2} = \frac{- ( - 2 ) - 2 3}{2 \cdot 1}

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

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

The solutions to the quadratic equation are:

y = 1 + 3, y = 1 - 3

(x + 1) 2 = x + 3

3 x^{2} - 23 x > 36

\frac{4 x - 5}{2} \leq \frac{1}{3}

7 x + 32 < 18

(\frac{6 ^{\frac{1}{5}}}{6 ^{\frac{2}{5}}})^{3}

What are the solutions to the equation y^2-2y-2=0 ?
The solutions to the equation y^2-2y-2=0 are y=1+sqrt(3),y=1-sqrt(3)
Find the zeros of y^2-2y-2=0
The zeros of y^2-2y-2=0 are y=1+sqrt(3),y=1-sqrt(3)