What are the solutions to the equation 2x^2-5x-2=0 ?

The solutions to the equation 2x^2-5x-2=0 are x=(5+sqrt(41))/4 ,x=(5-sqrt(41))/4

Find the zeros of 2x^2-5x-2=0

The zeros of 2x^2-5x-2=0 are x=(5+sqrt(41))/4 ,x=(5-sqrt(41))/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2 x^{2} - 5 x - 2 = 0

x = \frac{5 + 41}{4}, x = \frac{5 - 41}{4}

Decimal Notation

x = 2.85078 \dots, x = - 0.35078 \dots

Solution steps

2 x^{2} - 5 x - 2 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 2 ( - 2 )}{2 \cdot 2}

(- 5)^{2} - 4 \cdot 2 (- 2) = 41

x_{1, 2} = \frac{- ( - 5 ) \pm 41}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 41}{2 \cdot 2}, x_{2} = \frac{- ( - 5 ) - 41}{2 \cdot 2}

x = \frac{- ( - 5 ) + 41}{2 \cdot 2} : \frac{5 + 41}{4}

x = \frac{- ( - 5 ) - 41}{2 \cdot 2} : \frac{5 - 41}{4}

The solutions to the quadratic equation are:

x = \frac{5 + 41}{4}, x = \frac{5 - 41}{4}

\frac{9}{4}

(8 i)^{2}

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

2 (x - 4) + 3 x - x^{2} = 3

- 8 \leq 1 - 3 (x - 2) < 13

What are the solutions to the equation 2x^2-5x-2=0 ?
The solutions to the equation 2x^2-5x-2=0 are x=(5+sqrt(41))/4 ,x=(5-sqrt(41))/4
Find the zeros of 2x^2-5x-2=0
The zeros of 2x^2-5x-2=0 are x=(5+sqrt(41))/4 ,x=(5-sqrt(41))/4