What are the solutions to the equation x^2+4x+1=0 ?

The solutions to the equation x^2+4x+1=0 are x=-2+sqrt(3),x=-2-sqrt(3)

Find the zeros of x^2+4x+1=0

The zeros of x^2+4x+1=0 are x=-2+sqrt(3),x=-2-sqrt(3)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^{2} + 4 x + 1 = 0

x = - 2 + 3, x = - 2 - 3

Solution steps

x^{2} + 4 x + 1 = 0

Solve with the quadratic formula

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

4^{2} - 4 \cdot 1 \cdot 1 = 23

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

Separate the solutions

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

x = \frac{- 4 + 2 3}{2 \cdot 1} : - 2 + 3

x = \frac{- 4 - 2 3}{2 \cdot 1} : - 2 - 3

The solutions to the quadratic equation are:

x = - 2 + 3, x = - 2 - 3

x^{2} = - x

4 w^{2} + 16 w + 16

\frac{11}{6} - 1

- 6 x + 12 + 3 x = - 3 x + 6

\frac{6 ^{6}}{6 ^{1}}

What are the solutions to the equation x^2+4x+1=0 ?
The solutions to the equation x^2+4x+1=0 are x=-2+sqrt(3),x=-2-sqrt(3)
Find the zeros of x^2+4x+1=0
The zeros of x^2+4x+1=0 are x=-2+sqrt(3),x=-2-sqrt(3)