What are the solutions to the equation 3x^2+x=-2x-7 ?

The solutions to the equation 3x^2+x=-2x-7 are x=-1/2+i(5sqrt(3))/6 ,x=-1/2-i(5sqrt(3))/6

Find the zeros of 3x^2+x=-2x-7

The zeros of 3x^2+x=-2x-7 are x=-1/2+i(5sqrt(3))/6 ,x=-1/2-i(5sqrt(3))/6

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

3 x^{2} + x = - 2 x - 7

x = - \frac{1}{2} + i \frac{5 3}{6}, x = - \frac{1}{2} - i \frac{5 3}{6}

Solution steps

3 x^{2} + x = - 2 x - 7

Move

7

to the left side

3 x^{2} + x + 7 = - 2 x

Move

2 x

to the left side

3 x^{2} + 3 x + 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 3 \cdot 7}{2 \cdot 3}

Simplify

3^{2} - 4 \cdot 3 \cdot 7 : 53 i

x_{1, 2} = \frac{- 3 \pm 5 3 i}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 3 + 5 3 i}{2 \cdot 3}, x_{2} = \frac{- 3 - 5 3 i}{2 \cdot 3}

x = \frac{- 3 + 5 3 i}{2 \cdot 3} : - \frac{1}{2} + i \frac{5 3}{6}

x = \frac{- 3 - 5 3 i}{2 \cdot 3} : - \frac{1}{2} - i \frac{5 3}{6}

The solutions to the quadratic equation are:

x = - \frac{1}{2} + i \frac{5 3}{6}, x = - \frac{1}{2} - i \frac{5 3}{6}

\frac{a + 2}{4} \leq \frac{a - 1}{3}

\frac{7 + i}{7 - i}

45 - [4 - 2 y - 4 (y + 7)] = - 4 (1 + 3 y) - [4 - 3 (y + 2) - 2 (2 y - 5)]

- \frac{1}{2} (- 12 y + 16) = 5 y - 9

a^{2} \leq 1

What are the solutions to the equation 3x^2+x=-2x-7 ?
The solutions to the equation 3x^2+x=-2x-7 are x=-1/2+i(5sqrt(3))/6 ,x=-1/2-i(5sqrt(3))/6
Find the zeros of 3x^2+x=-2x-7
The zeros of 3x^2+x=-2x-7 are x=-1/2+i(5sqrt(3))/6 ,x=-1/2-i(5sqrt(3))/6