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

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

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

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

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

w = - 4, w = - 7

Solution steps

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

Solve with the quadratic formula

w_{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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

w = - 4, w = - 7

0 = - 16 t^{2} + 160 t

320 = - 20 x^{2} + 320 x - 780

t^{2} + 2 t = 15

1.8 \cdot 1 0^{- 5} = \frac{x ^{2}}{0.5}

x^{2} + 20 x - 100 = 0

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