What are the solutions to the equation y^2+14y+33=0 ?

The solutions to the equation y^2+14y+33=0 are y=-3,y=-11

Find the zeros of y^2+14y+33=0

The zeros of y^2+14y+33=0 are y=-3,y=-11

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y^{2} + 14 y + 33 = 0

y = - 3, y = - 11

Solution steps

y^{2} + 14 y + 33 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 14 \pm 1 4 ^{2} - 4 \cdot 1 \cdot 33}{2 \cdot 1}

1 4^{2} - 4 \cdot 1 \cdot 33 = 8

y_{1, 2} = \frac{- 14 \pm 8}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- 14 + 8}{2 \cdot 1}, y_{2} = \frac{- 14 - 8}{2 \cdot 1}

y = \frac{- 14 + 8}{2 \cdot 1} : - 3

y = \frac{- 14 - 8}{2 \cdot 1} : - 11

The solutions to the quadratic equation are:

y = - 3, y = - 11

x^{3} + 9 x

\frac{9 !}{2 ! ( 9 - 2 ) !}

4 b^{2} + 2 = 326

10 x + 18 < - 2

\frac{d ^{14}}{d ^{17}}

What are the solutions to the equation y^2+14y+33=0 ?
The solutions to the equation y^2+14y+33=0 are y=-3,y=-11
Find the zeros of y^2+14y+33=0
The zeros of y^2+14y+33=0 are y=-3,y=-11