What are the solutions to the equation -0.012x^2-10.8x+1200=0 ?

The solutions to the equation -0.012x^2-10.8x+1200=0 are x=100,x=-1000

Find the zeros of -0.012x^2-10.8x+1200=0

The zeros of -0.012x^2-10.8x+1200=0 are x=100,x=-1000

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

- 0.012 x^{2} - 10.8 x + 1200 = 0

x = 100, x = - 1000

Solution steps

- 0.012 x^{2} - 10.8 x + 1200 = 0

Multiply both sides by

1000

- 12 x^{2} - 10800 x + 1200000 = 0

Factor

- 12 x^{2} - 10800 x + 1200000 : - 12 (x - 100) (x + 1000)

- 12 (x - 100) (x + 1000) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 100 = 0 or x + 1000 = 0

Solve

x - 100 = 0 : x = 100

Solve

x + 1000 = 0 : x = - 1000

The solutions to the quadratic equation are:

x = 100, x = - 1000

- x^{2} - 25 x + 100 = 0

- 2 x (3 x - 2) = - 2

5 x - 2 x^{2} = 0

(x + 4) (2 x - 1) - 1 = - 3 x (x + 2)

(x - 5) (x + 1) = 2 (x - 2)^{2}

What are the solutions to the equation -0.012x^2-10.8x+1200=0 ?
The solutions to the equation -0.012x^2-10.8x+1200=0 are x=100,x=-1000
Find the zeros of -0.012x^2-10.8x+1200=0
The zeros of -0.012x^2-10.8x+1200=0 are x=100,x=-1000