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

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

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

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

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3 x^{2} + 12 x = 0

x = 0, x = - 4

Solution steps

3 x^{2} + 12 x = 0

Solve with the quadratic formula

x_{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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 0, x = - 4

155 \cdot \frac{π}{180}

(\frac{- 4}{5}) + \frac{1}{8}

(7 + x)^{2} + (7 - x)^{2} = 130

\frac{6}{4 + 5}

\frac{3 - 6}{3 + 6}

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