What are the solutions to the equation (y+3)^2=2y+6 ?

The solutions to the equation (y+3)^2=2y+6 are y=-1,y=-3

Find the zeros of (y+3)^2=2y+6

The zeros of (y+3)^2=2y+6 are y=-1,y=-3

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

y = - 1, y = - 3

Solution steps

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

Expand

(y + 3)^{2} : y^{2} + 6 y + 9

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

Move

6

to the left side

y^{2} + 6 y + 3 = 2 y

Move

2 y

to the left side

y^{2} + 4 y + 3 = 0

Solve with the quadratic formula

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

4^{2} - 4 \cdot 1 \cdot 3 = 2

y_{1, 2} = \frac{- 4 \pm 2}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- 4 + 2}{2 \cdot 1}, y_{2} = \frac{- 4 - 2}{2 \cdot 1}

y = \frac{- 4 + 2}{2 \cdot 1} : - 1

y = \frac{- 4 - 2}{2 \cdot 1} : - 3

The solutions to the quadratic equation are:

y = - 1, y = - 3

2 (6 - x) = 4 x

x^{3} + 4 x^{2} \leq 4 x + 16

2 a^{2} - 13 a - 16 = 8

12 x^{2} - 11 x - 15

5 (z + 3.41) = - 7.89 (2 z - 4) - 5.67

What are the solutions to the equation (y+3)^2=2y+6 ?
The solutions to the equation (y+3)^2=2y+6 are y=-1,y=-3
Find the zeros of (y+3)^2=2y+6
The zeros of (y+3)^2=2y+6 are y=-1,y=-3