What are the solutions to the equation 2y^2-10y-2=-12y ?

The solutions to the equation 2y^2-10y-2=-12y are y=(-1+sqrt(5))/2 ,y=-(1+sqrt(5))/2

Find the zeros of 2y^2-10y-2=-12y

The zeros of 2y^2-10y-2=-12y are y=(-1+sqrt(5))/2 ,y=-(1+sqrt(5))/2

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2 y^{2} - 10 y - 2 = - 12 y

y = \frac{- 1 + 5}{2}, y = - \frac{1 + 5}{2}

Decimal Notation

y = 0.61803 \dots, y = - 1.61803 \dots

Solution steps

2 y^{2} - 10 y - 2 = - 12 y

Move

12 y

to the left side

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

Solve with the quadratic formula

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

2^{2} - 4 \cdot 2 (- 2) = 25

y_{1, 2} = \frac{- 2 \pm 2 5}{2 \cdot 2}

Separate the solutions

y_{1} = \frac{- 2 + 2 5}{2 \cdot 2}, y_{2} = \frac{- 2 - 2 5}{2 \cdot 2}

y = \frac{- 2 + 2 5}{2 \cdot 2} : \frac{- 1 + 5}{2}

y = \frac{- 2 - 2 5}{2 \cdot 2} : - \frac{1 + 5}{2}

The solutions to the quadratic equation are:

y = \frac{- 1 + 5}{2}, y = - \frac{1 + 5}{2}

(x + 4)^{2} = 121

x \leq - 3 and x \geq - 2

\frac{1}{2} (0)

- x^{2} + 8 x - 15 < 0

(x + y)^{2} (x - y)^{2}

What are the solutions to the equation 2y^2-10y-2=-12y ?
The solutions to the equation 2y^2-10y-2=-12y are y=(-1+sqrt(5))/2 ,y=-(1+sqrt(5))/2
Find the zeros of 2y^2-10y-2=-12y
The zeros of 2y^2-10y-2=-12y are y=(-1+sqrt(5))/2 ,y=-(1+sqrt(5))/2