What are the solutions to the equation 12x^2+2x+52=0 ?

The solutions to the equation 12x^2+2x+52=0 are x=-1/12+i(sqrt(623))/(12),x=-1/12-i(sqrt(623))/(12)

Find the zeros of 12x^2+2x+52=0

The zeros of 12x^2+2x+52=0 are x=-1/12+i(sqrt(623))/(12),x=-1/12-i(sqrt(623))/(12)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

12 x^{2} + 2 x + 52 = 0

x = - \frac{1}{12} + i \frac{623}{12}, x = - \frac{1}{12} - i \frac{623}{12}

Solution steps

12 x^{2} + 2 x + 52 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 12 \cdot 52}{2 \cdot 12}

Simplify

2^{2} - 4 \cdot 12 \cdot 52 : 2623 i

x_{1, 2} = \frac{- 2 \pm 2 623 i}{2 \cdot 12}

Separate the solutions

x_{1} = \frac{- 2 + 2 623 i}{2 \cdot 12}, x_{2} = \frac{- 2 - 2 623 i}{2 \cdot 12}

x = \frac{- 2 + 2 623 i}{2 \cdot 12} : - \frac{1}{12} + i \frac{623}{12}

x = \frac{- 2 - 2 623 i}{2 \cdot 12} : - \frac{1}{12} - i \frac{623}{12}

The solutions to the quadratic equation are:

x = - \frac{1}{12} + i \frac{623}{12}, x = - \frac{1}{12} - i \frac{623}{12}

v, E k = \frac{1}{2} m v^{2}

x^{2} + 36 = 13 x

y^{2} + 7 y - 18 = 0

z^{2} - z + 5 = 0

v^{2} + 18 v + 81 = 50

What are the solutions to the equation 12x^2+2x+52=0 ?
The solutions to the equation 12x^2+2x+52=0 are x=-1/12+i(sqrt(623))/(12),x=-1/12-i(sqrt(623))/(12)
Find the zeros of 12x^2+2x+52=0
The zeros of 12x^2+2x+52=0 are x=-1/12+i(sqrt(623))/(12),x=-1/12-i(sqrt(623))/(12)