What are the solutions to the equation (y+1)(y+3)=-2 ?

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

Find the zeros of (y+1)(y+3)=-2

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

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(y + 1) (y + 3) = - 2

y = - 2 + i, y = - 2 - i

Solution steps

(y + 1) (y + 3) = - 2

Expand

(y + 1) (y + 3) : y^{2} + 4 y + 3

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

Move

2

to the left side

y^{2} + 4 y + 5 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot 5}{2 \cdot 1}

Simplify

4^{2} - 4 \cdot 1 \cdot 5 : 2 i

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

y = - 2 + i, y = - 2 - i

4 (3 x - 7)^{2} = 49

x^{2} + 6 x + 4 = 0

12 x^{2} - 52 x + 40 = 0

x, x^{2} + y^{2} - 4 x - 2 y + 25 = 0

2 x^{2} + 7 x = - 9

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