What are the solutions to the equation 2(y^2+3)=11-13y ?

The solutions to the equation 2(y^2+3)=11-13y are y=(-13+sqrt(209))/4 ,y=(-13-sqrt(209))/4

Find the zeros of 2(y^2+3)=11-13y

The zeros of 2(y^2+3)=11-13y are y=(-13+sqrt(209))/4 ,y=(-13-sqrt(209))/4

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2 (y^{2} + 3) = 11 - 13 y

y = \frac{- 13 + 209}{4}, y = \frac{- 13 - 209}{4}

Solution steps

2 (y^{2} + 3) = 11 - 13 y

Expand

2 (y^{2} + 3) : 2 y^{2} + 6

2 y^{2} + 6 = 11 - 13 y

Move

13 y

to the left side

2 y^{2} + 6 + 13 y = 11

Move

11

to the left side

2 y^{2} + 13 y - 5 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 13 \pm 1 3 ^{2} - 4 \cdot 2 ( - 5 )}{2 \cdot 2}

1 3^{2} - 4 \cdot 2 (- 5) = 209

y_{1, 2} = \frac{- 13 \pm 209}{2 \cdot 2}

Separate the solutions

y_{1} = \frac{- 13 + 209}{2 \cdot 2}, y_{2} = \frac{- 13 - 209}{2 \cdot 2}

y = \frac{- 13 + 209}{2 \cdot 2} : \frac{- 13 + 209}{4}

y = \frac{- 13 - 209}{2 \cdot 2} : \frac{- 13 - 209}{4}

The solutions to the quadratic equation are:

y = \frac{- 13 + 209}{4}, y = \frac{- 13 - 209}{4}

6 (1 - x) + 2 = 3 x^{2} - 17

d^{2} - 18 = 0

x^{2} + 4 x - 48 = 0

9 a^{2} - 40 = 0

5 x^{2} - 38 x - 16 = 0

What are the solutions to the equation 2(y^2+3)=11-13y ?
The solutions to the equation 2(y^2+3)=11-13y are y=(-13+sqrt(209))/4 ,y=(-13-sqrt(209))/4
Find the zeros of 2(y^2+3)=11-13y
The zeros of 2(y^2+3)=11-13y are y=(-13+sqrt(209))/4 ,y=(-13-sqrt(209))/4