What are the solutions to the equation y^2+12y+20=0 ?

The solutions to the equation y^2+12y+20=0 are y=-2,y=-10

Find the zeros of y^2+12y+20=0

The zeros of y^2+12y+20=0 are y=-2,y=-10

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y^{2} + 12 y + 20 = 0

y = - 2, y = - 10

Solution steps

y^{2} + 12 y + 20 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 1 \cdot 20}{2 \cdot 1}

1 2^{2} - 4 \cdot 1 \cdot 20 = 8

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

Separate the solutions

y_{1} = \frac{- 12 + 8}{2 \cdot 1}, y_{2} = \frac{- 12 - 8}{2 \cdot 1}

y = \frac{- 12 + 8}{2 \cdot 1} : - 2

y = \frac{- 12 - 8}{2 \cdot 1} : - 10

The solutions to the quadratic equation are:

y = - 2, y = - 10

3 = \frac{x}{2}

x^{2} + 14 x + 49 = 0

\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}

2 x^{2} + 7 x + 3 = 0

(- i)^{5}

What are the solutions to the equation y^2+12y+20=0 ?
The solutions to the equation y^2+12y+20=0 are y=-2,y=-10
Find the zeros of y^2+12y+20=0
The zeros of y^2+12y+20=0 are y=-2,y=-10