What are the solutions to the equation 3w^2+12w=0 ?

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

Find the zeros of 3w^2+12w=0

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

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3 w^{2} + 12 w = 0

w = 0, w = - 4

Solution steps

3 w^{2} + 12 w = 0

Solve with the quadratic formula

w_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 3 \cdot 0}{2 \cdot 3}

1 2^{2} - 4 \cdot 3 \cdot 0 = 12

w_{1, 2} = \frac{- 12 \pm 12}{2 \cdot 3}

Separate the solutions

w_{1} = \frac{- 12 + 12}{2 \cdot 3}, w_{2} = \frac{- 12 - 12}{2 \cdot 3}

w = \frac{- 12 + 12}{2 \cdot 3} : 0

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

The solutions to the quadratic equation are:

w = 0, w = - 4

3 x - 6 \leq - 15

10 \leq 2 x \leq x + 7

∣ 4 x - 7 ∣ \geq 3

\frac{1}{4} - \frac{3}{2}

i^{73}

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