What are the solutions to the equation 4w^2+28w=0 ?

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

Find the zeros of 4w^2+28w=0

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

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4 w^{2} + 28 w = 0

w = 0, w = - 7

Solution steps

4 w^{2} + 28 w = 0

Solve with the quadratic formula

w_{1, 2} = \frac{- 28 \pm 2 8 ^{2} - 4 \cdot 4 \cdot 0}{2 \cdot 4}

2 8^{2} - 4 \cdot 4 \cdot 0 = 28

w_{1, 2} = \frac{- 28 \pm 28}{2 \cdot 4}

Separate the solutions

w_{1} = \frac{- 28 + 28}{2 \cdot 4}, w_{2} = \frac{- 28 - 28}{2 \cdot 4}

w = \frac{- 28 + 28}{2 \cdot 4} : 0

w = \frac{- 28 - 28}{2 \cdot 4} : - 7

The solutions to the quadratic equation are:

w = 0, w = - 7

4 x^{2} + 12 x + 1 = 0

\frac{200}{1. 1 ^{5}}

\frac{1}{2} x > 3

x^{5} - 4 x^{3} + 8 x^{2} - 32

(4 x - 12) - 2 (4 x - 5) = (11 - 26) x

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