What are the solutions to the equation y^2+11y+28=0 ?

The solutions to the equation y^2+11y+28=0 are y=-4,y=-7

Find the zeros of y^2+11y+28=0

The zeros of y^2+11y+28=0 are y=-4,y=-7

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y^{2} + 11 y + 28 = 0

y = - 4, y = - 7

Solution steps

y^{2} + 11 y + 28 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 11 \pm 1 1 ^{2} - 4 \cdot 1 \cdot 28}{2 \cdot 1}

1 1^{2} - 4 \cdot 1 \cdot 28 = 3

y_{1, 2} = \frac{- 11 \pm 3}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- 11 + 3}{2 \cdot 1}, y_{2} = \frac{- 11 - 3}{2 \cdot 1}

y = \frac{- 11 + 3}{2 \cdot 1} : - 4

y = \frac{- 11 - 3}{2 \cdot 1} : - 7

The solutions to the quadratic equation are:

y = - 4, y = - 7

6 x^{2} - 23 x - 18 = 0

2 x^{2} + 5 x - 250 = 0

6 x^{2} + 2 x - 5 = 0

y^{2} - 32 y

\frac{( x - 2 ) ^{2}}{3} + \frac{14 x - 5}{6} = \frac{11}{6}

What are the solutions to the equation y^2+11y+28=0 ?
The solutions to the equation y^2+11y+28=0 are y=-4,y=-7
Find the zeros of y^2+11y+28=0
The zeros of y^2+11y+28=0 are y=-4,y=-7