What are the solutions to the equation 4w^2+12w=0 ?

The solutions to the equation 4w^2+12w=0 are w=0,w=-3

Find the zeros of 4w^2+12w=0

The zeros of 4w^2+12w=0 are w=0,w=-3

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4 w^{2} + 12 w = 0

w = 0, w = - 3

Solution steps

4 w^{2} + 12 w = 0

Solve with the quadratic formula

w_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 4 \cdot 0}{2 \cdot 4}

1 2^{2} - 4 \cdot 4 \cdot 0 = 12

w_{1, 2} = \frac{- 12 \pm 12}{2 \cdot 4}

Separate the solutions

w_{1} = \frac{- 12 + 12}{2 \cdot 4}, w_{2} = \frac{- 12 - 12}{2 \cdot 4}

w = \frac{- 12 + 12}{2 \cdot 4} : 0

w = \frac{- 12 - 12}{2 \cdot 4} : - 3

The solutions to the quadratic equation are:

w = 0, w = - 3

49 - x^{2} = 0

x^{2} + (\frac{2}{5})^{2} = 1

2 \leq x + 8 < 6

(18 + x) 2 = 30

\frac{1}{4} e^{2}

What are the solutions to the equation 4w^2+12w=0 ?
The solutions to the equation 4w^2+12w=0 are w=0,w=-3
Find the zeros of 4w^2+12w=0
The zeros of 4w^2+12w=0 are w=0,w=-3