What are the solutions to the equation 6x^2-9x+18=0 ?

The solutions to the equation 6x^2-9x+18=0 are x= 3/4+i(sqrt(39))/4 ,x= 3/4-i(sqrt(39))/4

Find the zeros of 6x^2-9x+18=0

The zeros of 6x^2-9x+18=0 are x= 3/4+i(sqrt(39))/4 ,x= 3/4-i(sqrt(39))/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

6 x^{2} - 9 x + 18 = 0

x = \frac{3}{4} + i \frac{39}{4}, x = \frac{3}{4} - i \frac{39}{4}

Solution steps

6 x^{2} - 9 x + 18 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 6 \cdot 18}{2 \cdot 6}

Simplify

(- 9)^{2} - 4 \cdot 6 \cdot 18 : 339 i

x_{1, 2} = \frac{- ( - 9 ) \pm 3 39 i}{2 \cdot 6}

Separate the solutions

x_{1} = \frac{- ( - 9 ) + 3 39 i}{2 \cdot 6}, x_{2} = \frac{- ( - 9 ) - 3 39 i}{2 \cdot 6}

x = \frac{- ( - 9 ) + 3 39 i}{2 \cdot 6} : \frac{3}{4} + i \frac{39}{4}

x = \frac{- ( - 9 ) - 3 39 i}{2 \cdot 6} : \frac{3}{4} - i \frac{39}{4}

The solutions to the quadratic equation are:

x = \frac{3}{4} + i \frac{39}{4}, x = \frac{3}{4} - i \frac{39}{4}

4 x^{2} + 12 x = - 9

7 (h - 2) = 56

\frac{x + 1}{x - 2} \geq \frac{x}{x - 1}

\frac{3}{16} + \frac{7}{16}

\frac{x + 9}{8} \geq 9

What are the solutions to the equation 6x^2-9x+18=0 ?
The solutions to the equation 6x^2-9x+18=0 are x= 3/4+i(sqrt(39))/4 ,x= 3/4-i(sqrt(39))/4
Find the zeros of 6x^2-9x+18=0
The zeros of 6x^2-9x+18=0 are x= 3/4+i(sqrt(39))/4 ,x= 3/4-i(sqrt(39))/4