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

The solutions to the equation y^2=-3y+9 are y=(-3+3sqrt(5))/2 ,y=(-3-3sqrt(5))/2

Find the zeros of y^2=-3y+9

The zeros of y^2=-3y+9 are y=(-3+3sqrt(5))/2 ,y=(-3-3sqrt(5))/2

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y^{2} = - 3 y + 9

y = \frac{- 3 + 3 5}{2}, y = \frac{- 3 - 3 5}{2}

Solution steps

y^{2} = - 3 y + 9

Move

9

to the left side

y^{2} - 9 = - 3 y

Move

3 y

to the left side

y^{2} - 9 + 3 y = 0

Write in the standard form

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

y^{2} + 3 y - 9 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot ( - 9 )}{2 \cdot 1}

3^{2} - 4 \cdot 1 \cdot (- 9) = 35

y_{1, 2} = \frac{- 3 \pm 3 5}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- 3 + 3 5}{2 \cdot 1}, y_{2} = \frac{- 3 - 3 5}{2 \cdot 1}

y = \frac{- 3 + 3 5}{2 \cdot 1} : \frac{- 3 + 3 5}{2}

y = \frac{- 3 - 3 5}{2 \cdot 1} : \frac{- 3 - 3 5}{2}

The solutions to the quadratic equation are:

y = \frac{- 3 + 3 5}{2}, y = \frac{- 3 - 3 5}{2}

7 x + 10 - 9 - 9 x = 4 x + 7 - 6 - 6 x

0 = (x - 2)^{2} - 1

121 y^{2} - 144

20 x^{2} + 28

- 3 + 4

What are the solutions to the equation y^2=-3y+9 ?
The solutions to the equation y^2=-3y+9 are y=(-3+3sqrt(5))/2 ,y=(-3-3sqrt(5))/2
Find the zeros of y^2=-3y+9
The zeros of y^2=-3y+9 are y=(-3+3sqrt(5))/2 ,y=(-3-3sqrt(5))/2