What are the solutions to the equation 2(y^2+3y)=9 ?

The solutions to the equation 2(y^2+3y)=9 are y=(3(sqrt(3)-1))/2 ,y=-(3(1+sqrt(3)))/2

Find the zeros of 2(y^2+3y)=9

The zeros of 2(y^2+3y)=9 are y=(3(sqrt(3)-1))/2 ,y=-(3(1+sqrt(3)))/2

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2 (y^{2} + 3 y) = 9

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

Solution steps

2 (y^{2} + 3 y) = 9

Expand

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

2 y^{2} + 6 y = 9

Move

9

to the left side

2 y^{2} + 6 y - 9 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 2 ( - 9 )}{2 \cdot 2}

6^{2} - 4 \cdot 2 (- 9) = 63

y_{1, 2} = \frac{- 6 \pm 6 3}{2 \cdot 2}

Separate the solutions

y_{1} = \frac{- 6 + 6 3}{2 \cdot 2}, y_{2} = \frac{- 6 - 6 3}{2 \cdot 2}

y = \frac{- 6 + 6 3}{2 \cdot 2} : \frac{3 ( 3 - 1 )}{2}

y = \frac{- 6 - 6 3}{2 \cdot 2} : - \frac{3 ( 1 + 3 )}{2}

The solutions to the quadratic equation are:

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

3 x^{2} - 110 x + 900 = 0

(x + 4)^{2} = - 36

x^{2} + (b + 4) x = 0

x^{2} + 4 x = 5

2 x^{2} + 7 = 31

What are the solutions to the equation 2(y^2+3y)=9 ?
The solutions to the equation 2(y^2+3y)=9 are y=(3(sqrt(3)-1))/2 ,y=-(3(1+sqrt(3)))/2
Find the zeros of 2(y^2+3y)=9
The zeros of 2(y^2+3y)=9 are y=(3(sqrt(3)-1))/2 ,y=-(3(1+sqrt(3)))/2