What are the solutions to the equation 3x^2+2x+6=5 ?

The solutions to the equation 3x^2+2x+6=5 are x=-1/3+i(sqrt(2))/3 ,x=-1/3-i(sqrt(2))/3

Find the zeros of 3x^2+2x+6=5

The zeros of 3x^2+2x+6=5 are x=-1/3+i(sqrt(2))/3 ,x=-1/3-i(sqrt(2))/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

3 x^{2} + 2 x + 6 = 5

x = - \frac{1}{3} + i \frac{2}{3}, x = - \frac{1}{3} - i \frac{2}{3}

Solution steps

3 x^{2} + 2 x + 6 = 5

Move

5

to the left side

3 x^{2} + 2 x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 3 \cdot 1}{2 \cdot 3}

Simplify

2^{2} - 4 \cdot 3 \cdot 1 : 22 i

x_{1, 2} = \frac{- 2 \pm 2 2 i}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 2 + 2 2 i}{2 \cdot 3}, x_{2} = \frac{- 2 - 2 2 i}{2 \cdot 3}

x = \frac{- 2 + 2 2 i}{2 \cdot 3} : - \frac{1}{3} + i \frac{2}{3}

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

The solutions to the quadratic equation are:

x = - \frac{1}{3} + i \frac{2}{3}, x = - \frac{1}{3} - i \frac{2}{3}

2 + x < 9 x + 6

2 (\frac{1}{3})^{2} + 5 (\frac{1}{3}) + 6

6 + 3 ∣ 2 x + 7 ∣ = 30

\frac{8}{9} + 1

\frac{( n + 1 ) !}{( n + 2 ) !}

What are the solutions to the equation 3x^2+2x+6=5 ?
The solutions to the equation 3x^2+2x+6=5 are x=-1/3+i(sqrt(2))/3 ,x=-1/3-i(sqrt(2))/3
Find the zeros of 3x^2+2x+6=5
The zeros of 3x^2+2x+6=5 are x=-1/3+i(sqrt(2))/3 ,x=-1/3-i(sqrt(2))/3