What are the solutions to the equation (2x+1)(x-2)=5 ?

The solutions to the equation (2x+1)(x-2)=5 are x=(3+sqrt(65))/4 ,x=(3-sqrt(65))/4

Find the zeros of (2x+1)(x-2)=5

The zeros of (2x+1)(x-2)=5 are x=(3+sqrt(65))/4 ,x=(3-sqrt(65))/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

x = \frac{3 + 65}{4}, x = \frac{3 - 65}{4}

Decimal Notation

x = 2.76556 \dots, x = - 1.26556 \dots

Solution steps

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

Expand

(2 x + 1) (x - 2) : 2 x^{2} - 3 x - 2

2 x^{2} - 3 x - 2 = 5

Move

5

to the left side

2 x^{2} - 3 x - 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 2 ( - 7 )}{2 \cdot 2}

(- 3)^{2} - 4 \cdot 2 (- 7) = 65

x_{1, 2} = \frac{- ( - 3 ) \pm 65}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 65}{2 \cdot 2}, x_{2} = \frac{- ( - 3 ) - 65}{2 \cdot 2}

x = \frac{- ( - 3 ) + 65}{2 \cdot 2} : \frac{3 + 65}{4}

x = \frac{- ( - 3 ) - 65}{2 \cdot 2} : \frac{3 - 65}{4}

The solutions to the quadratic equation are:

x = \frac{3 + 65}{4}, x = \frac{3 - 65}{4}

x, x^{2} + y^{2} + 18 = 0

8 = 3 x^{2} - 2 x

2 x (x + 2) = 0

(x + 5)^{2} = 29

2 x^{2} + 5 x - 3 = x^{2} - x + 13

What are the solutions to the equation (2x+1)(x-2)=5 ?
The solutions to the equation (2x+1)(x-2)=5 are x=(3+sqrt(65))/4 ,x=(3-sqrt(65))/4
Find the zeros of (2x+1)(x-2)=5
The zeros of (2x+1)(x-2)=5 are x=(3+sqrt(65))/4 ,x=(3-sqrt(65))/4