What are the solutions to the equation y^2+7y+12=0 ?

The solutions to the equation y^2+7y+12=0 are y=-3,y=-4

Find the zeros of y^2+7y+12=0

The zeros of y^2+7y+12=0 are y=-3,y=-4

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y^{2} + 7 y + 12 = 0

y = - 3, y = - 4

Solution steps

y^{2} + 7 y + 12 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 1 \cdot 12}{2 \cdot 1}

7^{2} - 4 \cdot 1 \cdot 12 = 1

y_{1, 2} = \frac{- 7 \pm 1}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- 7 + 1}{2 \cdot 1}, y_{2} = \frac{- 7 - 1}{2 \cdot 1}

y = \frac{- 7 + 1}{2 \cdot 1} : - 3

y = \frac{- 7 - 1}{2 \cdot 1} : - 4

The solutions to the quadratic equation are:

y = - 3, y = - 4

x + 9 = 2 (x - 1)^{2}

1 < 1 - \frac{1}{6} x < 4

15 + 7 x \geq 2 x - 5

2 x + 80 = 180

2 - 3 m < \frac{1}{2}

What are the solutions to the equation y^2+7y+12=0 ?
The solutions to the equation y^2+7y+12=0 are y=-3,y=-4
Find the zeros of y^2+7y+12=0
The zeros of y^2+7y+12=0 are y=-3,y=-4