What are the solutions to the equation 7x^2-2x+3=0 ?

The solutions to the equation 7x^2-2x+3=0 are x= 1/7+i(2sqrt(5))/7 ,x= 1/7-i(2sqrt(5))/7

Find the zeros of 7x^2-2x+3=0

The zeros of 7x^2-2x+3=0 are x= 1/7+i(2sqrt(5))/7 ,x= 1/7-i(2sqrt(5))/7

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

x = \frac{1}{7} + i \frac{2 5}{7}, x = \frac{1}{7} - i \frac{2 5}{7}

Solution steps

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

Solve with the quadratic formula

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

Simplify

(- 2)^{2} - 4 \cdot 7 \cdot 3 : 45 i

x_{1, 2} = \frac{- ( - 2 ) \pm 4 5 i}{2 \cdot 7}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 4 5 i}{2 \cdot 7}, x_{2} = \frac{- ( - 2 ) - 4 5 i}{2 \cdot 7}

x = \frac{- ( - 2 ) + 4 5 i}{2 \cdot 7} : \frac{1}{7} + i \frac{2 5}{7}

x = \frac{- ( - 2 ) - 4 5 i}{2 \cdot 7} : \frac{1}{7} - i \frac{2 5}{7}

The solutions to the quadratic equation are:

x = \frac{1}{7} + i \frac{2 5}{7}, x = \frac{1}{7} - i \frac{2 5}{7}

\frac{- 3 - 2}{3 17}

6 x^{2} + 21 x - 12

\frac{x ^{- 2} - y ^{- 2}}{x ^{- 1} + y ^{- 1}}

x y - y^{2}

6 - 2 t^{2} = 9 t + 15

What are the solutions to the equation 7x^2-2x+3=0 ?
The solutions to the equation 7x^2-2x+3=0 are x= 1/7+i(2sqrt(5))/7 ,x= 1/7-i(2sqrt(5))/7
Find the zeros of 7x^2-2x+3=0
The zeros of 7x^2-2x+3=0 are x= 1/7+i(2sqrt(5))/7 ,x= 1/7-i(2sqrt(5))/7