What are the solutions to the equation 5x^2+x+4=0 ?

The solutions to the equation 5x^2+x+4=0 are x=-1/10+i(sqrt(79))/(10),x=-1/10-i(sqrt(79))/(10)

Find the zeros of 5x^2+x+4=0

The zeros of 5x^2+x+4=0 are x=-1/10+i(sqrt(79))/(10),x=-1/10-i(sqrt(79))/(10)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

5 x^{2} + x + 4 = 0

x = - \frac{1}{10} + i \frac{79}{10}, x = - \frac{1}{10} - i \frac{79}{10}

Solution steps

5 x^{2} + x + 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 5 \cdot 4}{2 \cdot 5}

Simplify

1^{2} - 4 \cdot 5 \cdot 4 : 79 i

x_{1, 2} = \frac{- 1 \pm 79 i}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- 1 + 79 i}{2 \cdot 5}, x_{2} = \frac{- 1 - 79 i}{2 \cdot 5}

x = \frac{- 1 + 79 i}{2 \cdot 5} : - \frac{1}{10} + i \frac{79}{10}

x = \frac{- 1 - 79 i}{2 \cdot 5} : - \frac{1}{10} - i \frac{79}{10}

The solutions to the quadratic equation are:

x = - \frac{1}{10} + i \frac{79}{10}, x = - \frac{1}{10} - i \frac{79}{10}

(2 x)^{- 3}

\frac{3}{5} + \frac{3}{2}

0 = 144 t - 16 t^{2}

0.2

3 (x - 2) 2 = 18

What are the solutions to the equation 5x^2+x+4=0 ?
The solutions to the equation 5x^2+x+4=0 are x=-1/10+i(sqrt(79))/(10),x=-1/10-i(sqrt(79))/(10)
Find the zeros of 5x^2+x+4=0
The zeros of 5x^2+x+4=0 are x=-1/10+i(sqrt(79))/(10),x=-1/10-i(sqrt(79))/(10)