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

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

Find the zeros of x^2=-1-4x

The zeros of x^2=-1-4x 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} = - 1 - 4 x

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

Decimal Notation

x = - 0.26794 \dots, x = - 3.73205 \dots

Solution steps

x^{2} = - 1 - 4 x

Move

4 x

to the left side

x^{2} + 4 x = - 1

Move

1

to the left side

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

3 x^{2} - 15 = 4 x

2 x^{2} - 10 x = - 7 x - 2

x^{2} - 18 x - 63 = 0

b^{2} = 81

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

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