What are the solutions to the equation y^2+15y+36=0 ?

The solutions to the equation y^2+15y+36=0 are y=-3,y=-12

Find the zeros of y^2+15y+36=0

The zeros of y^2+15y+36=0 are y=-3,y=-12

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y^{2} + 15 y + 36 = 0

y = - 3, y = - 12

Solution steps

y^{2} + 15 y + 36 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 15 \pm 1 5 ^{2} - 4 \cdot 1 \cdot 36}{2 \cdot 1}

1 5^{2} - 4 \cdot 1 \cdot 36 = 9

y_{1, 2} = \frac{- 15 \pm 9}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- 15 + 9}{2 \cdot 1}, y_{2} = \frac{- 15 - 9}{2 \cdot 1}

y = \frac{- 15 + 9}{2 \cdot 1} : - 3

y = \frac{- 15 - 9}{2 \cdot 1} : - 12

The solutions to the quadratic equation are:

y = - 3, y = - 12

2 x^{2} + 6 x - 14 = 0

2 x^{2} - x - 6 = - 3 x - 2

4 - x^{2}

3 x^{2} + 10 x - 6 = 0

- x^{2} + 6 x + 4

What are the solutions to the equation y^2+15y+36=0 ?
The solutions to the equation y^2+15y+36=0 are y=-3,y=-12
Find the zeros of y^2+15y+36=0
The zeros of y^2+15y+36=0 are y=-3,y=-12