What are the solutions to the equation 2x^2-8x-32=0 ?

The solutions to the equation 2x^2-8x-32=0 are x=2(1+sqrt(5)),x=2(1-sqrt(5))

Find the zeros of 2x^2-8x-32=0

The zeros of 2x^2-8x-32=0 are x=2(1+sqrt(5)),x=2(1-sqrt(5))

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 - 32 = 0

x = 2 (1 + 5), x = 2 (1 - 5)

Decimal Notation

x = 6.47213 \dots, x = - 2.47213 \dots

Solution steps

2 x^{2} - 8 x - 32 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 2 ( - 32 )}{2 \cdot 2}

(- 8)^{2} - 4 \cdot 2 (- 32) = 85

x_{1, 2} = \frac{- ( - 8 ) \pm 8 5}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 8 ) + 8 5}{2 \cdot 2}, x_{2} = \frac{- ( - 8 ) - 8 5}{2 \cdot 2}

x = \frac{- ( - 8 ) + 8 5}{2 \cdot 2} : 2 (1 + 5)

x = \frac{- ( - 8 ) - 8 5}{2 \cdot 2} : 2 (1 - 5)

The solutions to the quadratic equation are:

x = 2 (1 + 5), x = 2 (1 - 5)

\frac{35 x ^{3} y z ^{2}}{21 x ^{2} y ^{3} z}

\frac{- 6 ( - 4 ) + ( - 2 ) ^{3}}{- 4}

8 x^{2} = - 26 x

x^{2} + (x + 7)^{2} = (2 x - 7)^{2}

3 x^{2} + 13 x + 4

What are the solutions to the equation 2x^2-8x-32=0 ?
The solutions to the equation 2x^2-8x-32=0 are x=2(1+sqrt(5)),x=2(1-sqrt(5))
Find the zeros of 2x^2-8x-32=0
The zeros of 2x^2-8x-32=0 are x=2(1+sqrt(5)),x=2(1-sqrt(5))