What are the solutions to the equation y^2+15y+54=0 ?

The solutions to the equation y^2+15y+54=0 are y=-6,y=-9

Find the zeros of y^2+15y+54=0

The zeros of y^2+15y+54=0 are y=-6,y=-9

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y^{2} + 15 y + 54 = 0

y = - 6, y = - 9

Solution steps

y^{2} + 15 y + 54 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 15 \pm 1 5 ^{2} - 4 \cdot 1 \cdot 54}{2 \cdot 1}

1 5^{2} - 4 \cdot 1 \cdot 54 = 3

y_{1, 2} = \frac{- 15 \pm 3}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- 15 + 3}{2 \cdot 1}, y_{2} = \frac{- 15 - 3}{2 \cdot 1}

y = \frac{- 15 + 3}{2 \cdot 1} : - 6

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

The solutions to the quadratic equation are:

y = - 6, y = - 9

6 x^{2} + 38 x + 60 = 10

x^{2} + (- \frac{7}{25})^{2} = 1

51 + \frac{769}{32} \cdot x - \frac{1}{2} \cdot 10 \cdot (x)^{2} = 0

3 X^{2} - 27 = 0

(x^{2} + 6) = 0

What are the solutions to the equation y^2+15y+54=0 ?
The solutions to the equation y^2+15y+54=0 are y=-6,y=-9
Find the zeros of y^2+15y+54=0
The zeros of y^2+15y+54=0 are y=-6,y=-9