What are the solutions to the equation 2y^2+9=-(9y+1/5 y) ?

The solutions to the equation 2y^2+9=-(9y+1/5 y) are y=(-23+sqrt(79))/(10),y=-(23+sqrt(79))/(10)

Find the zeros of 2y^2+9=-(9y+1/5 y)

The zeros of 2y^2+9=-(9y+1/5 y) are y=(-23+sqrt(79))/(10),y=-(23+sqrt(79))/(10)

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2 y^{2} + 9 = - (9 y + \frac{1}{5} y)

y = \frac{- 23 + 79}{10}, y = - \frac{23 + 79}{10}

Decimal Notation

y = - 1.41118 \dots, y = - 3.18881 \dots

Solution steps

2 y^{2} + 9 = - (9 y + \frac{1}{5} y)

Expand

- (9 y + \frac{1}{5} y) : - \frac{46}{5} y

2 y^{2} + 9 = - \frac{46}{5} y

Move

\frac{46}{5} y

to the left side

2 y^{2} + 9 + \frac{46}{5} y = 0

Write in the standard form

a x^{2} + b x + c = 0

2 y^{2} + \frac{46 y}{5} + 9 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- \frac{46}{5} \pm ( \frac{46}{5} ) ^{2} - 4 \cdot 2 \cdot 9}{2 \cdot 2}

(\frac{46}{5})^{2} - 4 \cdot 2 \cdot 9 = \frac{2 79}{5}

y_{1, 2} = \frac{- \frac{46}{5} \pm \frac{2 79}{5}}{2 \cdot 2}

Separate the solutions

y_{1} = \frac{- \frac{46}{5} + \frac{2 79}{5}}{2 \cdot 2}, y_{2} = \frac{- \frac{46}{5} - \frac{2 79}{5}}{2 \cdot 2}

y = \frac{- \frac{46}{5} + \frac{2 79}{5}}{2 \cdot 2} : \frac{- 23 + 79}{10}

y = \frac{- \frac{46}{5} - \frac{2 79}{5}}{2 \cdot 2} : - \frac{23 + 79}{10}

The solutions to the quadratic equation are:

y = \frac{- 23 + 79}{10}, y = - \frac{23 + 79}{10}

- 4 x^{2} + 7 x + 3 = 0

2 x^{2} - 4 x - 3

7 x^{2} = 343

λ^{2} - 2 λ + 2 = 0

144 r^{2} + 15 = 51

What are the solutions to the equation 2y^2+9=-(9y+1/5 y) ?
The solutions to the equation 2y^2+9=-(9y+1/5 y) are y=(-23+sqrt(79))/(10),y=-(23+sqrt(79))/(10)
Find the zeros of 2y^2+9=-(9y+1/5 y)
The zeros of 2y^2+9=-(9y+1/5 y) are y=(-23+sqrt(79))/(10),y=-(23+sqrt(79))/(10)