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

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

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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

x = 1 + 3, x = 1 - 3

Decimal Notation

x = 2.73205 \dots, x = - 0.73205 \dots

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 1 + 3, x = 1 - 3

t^{2} + 5 t - 24 = 0

3 w + 5 = 23

n + 1 \leq 5

- 3^{3} x^{- 5} + 5 x^{2} y^{2} \cdot \frac{4 ^{2} x ^{- 7}}{y ^{2}}

(5^{x})^{2}

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