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

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

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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

x = 0, x = 3

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 0, x = 3

3 x^{2} - 8 x - 2 = 0

(x - 4) (x + 6) = 0

3 n^{2} = 27

x^{2} + 22 x = 0

x^{2} + 6 x + 9 = 81

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