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

The solutions to the equation 12x^2+4x=0 are x=0,x=-1/3

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

The zeros of 12x^2+4x=0 are x=0,x=-1/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

12 x^{2} + 4 x = 0

x = 0, x = - \frac{1}{3}

Solution steps

12 x^{2} + 4 x = 0

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

x = \frac{- 4 - 4}{2 \cdot 12} : - \frac{1}{3}

The solutions to the quadratic equation are:

x = 0, x = - \frac{1}{3}

12 x^{2} - 252 x + 902 = 0

7 m^{2} + 40 = 43 m

(x - 7)^{2} = 6

(x^{2} - 4 x + 4) = 0

2 x^{2} - 20 x = 0

What are the solutions to the equation 12x^2+4x=0 ?
The solutions to the equation 12x^2+4x=0 are x=0,x=-1/3
Find the zeros of 12x^2+4x=0
The zeros of 12x^2+4x=0 are x=0,x=-1/3