What are the solutions to the equation 12x^2+14x=-5 ?

The solutions to the equation 12x^2+14x=-5 are x=-7/12+i(sqrt(11))/(12),x=-7/12-i(sqrt(11))/(12)

Find the zeros of 12x^2+14x=-5

The zeros of 12x^2+14x=-5 are x=-7/12+i(sqrt(11))/(12),x=-7/12-i(sqrt(11))/(12)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

12 x^{2} + 14 x = - 5

x = - \frac{7}{12} + i \frac{11}{12}, x = - \frac{7}{12} - i \frac{11}{12}

Solution steps

12 x^{2} + 14 x = - 5

Move

5

to the left side

12 x^{2} + 14 x + 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 14 \pm 1 4 ^{2} - 4 \cdot 12 \cdot 5}{2 \cdot 12}

Simplify

1 4^{2} - 4 \cdot 12 \cdot 5 : 211 i

x_{1, 2} = \frac{- 14 \pm 2 11 i}{2 \cdot 12}

Separate the solutions

x_{1} = \frac{- 14 + 2 11 i}{2 \cdot 12}, x_{2} = \frac{- 14 - 2 11 i}{2 \cdot 12}

x = \frac{- 14 + 2 11 i}{2 \cdot 12} : - \frac{7}{12} + i \frac{11}{12}

x = \frac{- 14 - 2 11 i}{2 \cdot 12} : - \frac{7}{12} - i \frac{11}{12}

The solutions to the quadratic equation are:

x = - \frac{7}{12} + i \frac{11}{12}, x = - \frac{7}{12} - i \frac{11}{12}

1 - x = x^{2}

- 7 m^{2} = - 4 - 4 m - 8 m^{2}

(- 14 - x) (14 - x) + 192 = 0

(x + 4)^{2} - 36 = 0

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

What are the solutions to the equation 12x^2+14x=-5 ?
The solutions to the equation 12x^2+14x=-5 are x=-7/12+i(sqrt(11))/(12),x=-7/12-i(sqrt(11))/(12)
Find the zeros of 12x^2+14x=-5
The zeros of 12x^2+14x=-5 are x=-7/12+i(sqrt(11))/(12),x=-7/12-i(sqrt(11))/(12)