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

The solutions to the equation 2x^2+x-2=0 are x=(-1+sqrt(17))/4 ,x=(-1-sqrt(17))/4

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

The zeros of 2x^2+x-2=0 are x=(-1+sqrt(17))/4 ,x=(-1-sqrt(17))/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

x = \frac{- 1 + 17}{4}, x = \frac{- 1 - 17}{4}

Decimal Notation

x = 0.78077 \dots, x = - 1.28077 \dots

Solution steps

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

Solve with the quadratic formula

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

1^{2} - 4 \cdot 2 (- 2) = 17

x_{1, 2} = \frac{- 1 \pm 17}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 1 + 17}{2 \cdot 2}, x_{2} = \frac{- 1 - 17}{2 \cdot 2}

x = \frac{- 1 + 17}{2 \cdot 2} : \frac{- 1 + 17}{4}

x = \frac{- 1 - 17}{2 \cdot 2} : \frac{- 1 - 17}{4}

The solutions to the quadratic equation are:

x = \frac{- 1 + 17}{4}, x = \frac{- 1 - 17}{4}

\frac{\frac{2 x - 6}{y}}{\frac{x - 3}{y + 1}}

(0.008)^{- \frac{2}{3}}

x^{7} - 2 x^{5} - 1

\frac{t ^{2}}{2} + 4 t + 5 = 0

(- 2 + - 8) + (5 - - 50)

What are the solutions to the equation 2x^2+x-2=0 ?
The solutions to the equation 2x^2+x-2=0 are x=(-1+sqrt(17))/4 ,x=(-1-sqrt(17))/4
Find the zeros of 2x^2+x-2=0
The zeros of 2x^2+x-2=0 are x=(-1+sqrt(17))/4 ,x=(-1-sqrt(17))/4