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

The solutions to the equation 2x^2-8x-4=0 are x=2+sqrt(6),x=2-sqrt(6)

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

The zeros of 2x^2-8x-4=0 are x=2+sqrt(6),x=2-sqrt(6)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

x = 2 + 6, x = 2 - 6

Decimal Notation

x = 4.44948 \dots, x = - 0.44948 \dots

Solution steps

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

Solve with the quadratic formula

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

(- 8)^{2} - 4 \cdot 2 (- 4) = 46

x_{1, 2} = \frac{- ( - 8 ) \pm 4 6}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 8 ) + 4 6}{2 \cdot 2}, x_{2} = \frac{- ( - 8 ) - 4 6}{2 \cdot 2}

x = \frac{- ( - 8 ) + 4 6}{2 \cdot 2} : 2 + 6

x = \frac{- ( - 8 ) - 4 6}{2 \cdot 2} : 2 - 6

The solutions to the quadratic equation are:

x = 2 + 6, x = 2 - 6

8 x^{2} - 10 x - 3 = 0

v^{2} + (- \frac{2}{3} + v)^{2} = 1

15 x - 2 = 85 - 4 (x - 2)

\frac{2}{7}

1 < - 6 x - 17

What are the solutions to the equation 2x^2-8x-4=0 ?
The solutions to the equation 2x^2-8x-4=0 are x=2+sqrt(6),x=2-sqrt(6)
Find the zeros of 2x^2-8x-4=0
The zeros of 2x^2-8x-4=0 are x=2+sqrt(6),x=2-sqrt(6)