What are the solutions to the equation y^2+8y+12=0 ?

The solutions to the equation y^2+8y+12=0 are y=-2,y=-6

Find the zeros of y^2+8y+12=0

The zeros of y^2+8y+12=0 are y=-2,y=-6

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y^{2} + 8 y + 12 = 0

y = - 2, y = - 6

Solution steps

y^{2} + 8 y + 12 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 1 \cdot 12}{2 \cdot 1}

8^{2} - 4 \cdot 1 \cdot 12 = 4

y_{1, 2} = \frac{- 8 \pm 4}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- 8 + 4}{2 \cdot 1}, y_{2} = \frac{- 8 - 4}{2 \cdot 1}

y = \frac{- 8 + 4}{2 \cdot 1} : - 2

y = \frac{- 8 - 4}{2 \cdot 1} : - 6

The solutions to the quadratic equation are:

y = - 2, y = - 6

z^{2} + 2 z + 3 = 0

x^{2} + 6 x + 45 = 0

- 3 x + 0 = - 4

What are the solutions to the equation y^2+8y+12=0 ?
The solutions to the equation y^2+8y+12=0 are y=-2,y=-6
Find the zeros of y^2+8y+12=0
The zeros of y^2+8y+12=0 are y=-2,y=-6