What are the solutions to the equation solvefor y,y^2+6y+9=x+8 ?

The solutions to the equation solvefor y,y^2+6y+9=x+8 are y=-3+sqrt(x+8),y=-sqrt(x+8)-3

Find the zeros of solvefor y,y^2+6y+9=x+8

The zeros of solvefor y,y^2+6y+9=x+8 are y=-3+sqrt(x+8),y=-sqrt(x+8)-3

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solve for

y, y^{2} + 6 y + 9 = x + 8

y = - 3 + x + 8, y = - x + 8 - 3

Solution steps

y^{2} + 6 y + 9 = x + 8

Move

8

to the left side

y^{2} + 6 y + 1 = x

Move

x

to the left side

y^{2} + 6 y + 1 - x = 0

Solve with the quadratic formula

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

Simplify

6^{2} - 4 \cdot 1 \cdot (1 - x) : 2 x + 8

y_{1, 2} = \frac{- 6 \pm 2 x + 8}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- 6 + 2 x + 8}{2 \cdot 1}, y_{2} = \frac{- 6 - 2 x + 8}{2 \cdot 1}

y = \frac{- 6 + 2 x + 8}{2 \cdot 1} : - 3 + x + 8

y = \frac{- 6 - 2 x + 8}{2 \cdot 1} : - x + 8 - 3

The solutions to the quadratic equation are:

y = - 3 + x + 8, y = - x + 8 - 3

3 x^{2} - 5 (2 x + 7) = 0

3 b^{2} - b + 5 = 6

x, (2 x - 9 y)^{2} + 7 y = 27 x

3 y^{2} + 2 = 5 y

\frac{( 2 x + 3 ) ^{2}}{2} = - 1

What are the solutions to the equation solvefor y,y^2+6y+9=x+8 ?
The solutions to the equation solvefor y,y^2+6y+9=x+8 are y=-3+sqrt(x+8),y=-sqrt(x+8)-3
Find the zeros of solvefor y,y^2+6y+9=x+8
The zeros of solvefor y,y^2+6y+9=x+8 are y=-3+sqrt(x+8),y=-sqrt(x+8)-3