What are the solutions to the equation 0=x^2+4x-8 ?

The solutions to the equation 0=x^2+4x-8 are x=2(sqrt(3)-1),x=-2(1+sqrt(3))

Find the zeros of 0=x^2+4x-8

The zeros of 0=x^2+4x-8 are x=2(sqrt(3)-1),x=-2(1+sqrt(3))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

0 = x^{2} + 4 x - 8

x = 2 (3 - 1), x = - 2 (1 + 3)

Decimal Notation

x = 1.46410 \dots, x = - 5.46410 \dots

Solution steps

0 = x^{2} + 4 x - 8

Switch sides

x^{2} + 4 x - 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot ( - 8 )}{2 \cdot 1}

4^{2} - 4 \cdot 1 \cdot (- 8) = 43

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 2 (3 - 1), x = - 2 (1 + 3)

\frac{n}{4} = 4

6 x^{4} + 4 x^{3} - 9 x^{2} - 6 x

\frac{2 x - 6}{3} = - 12

2 x^{2} + x + 1 = 64 + 40 + 2

\frac{π}{\frac{1}{4}}

What are the solutions to the equation 0=x^2+4x-8 ?
The solutions to the equation 0=x^2+4x-8 are x=2(sqrt(3)-1),x=-2(1+sqrt(3))
Find the zeros of 0=x^2+4x-8
The zeros of 0=x^2+4x-8 are x=2(sqrt(3)-1),x=-2(1+sqrt(3))