What are the solutions to the equation x^2-x-1=9 ?

The solutions to the equation x^2-x-1=9 are x=(1+sqrt(41))/2 ,x=(1-sqrt(41))/2

Find the zeros of x^2-x-1=9

The zeros of x^2-x-1=9 are x=(1+sqrt(41))/2 ,x=(1-sqrt(41))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^{2} - x - 1 = 9

x = \frac{1 + 41}{2}, x = \frac{1 - 41}{2}

Decimal Notation

x = 3.70156 \dots, x = - 2.70156 \dots

Solution steps

x^{2} - x - 1 = 9

Move

9

to the left side

x^{2} - x - 10 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot ( - 10 )}{2 \cdot 1}

(- 1)^{2} - 4 \cdot 1 \cdot (- 10) = 41

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

Separate the solutions

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

x = \frac{- ( - 1 ) + 41}{2 \cdot 1} : \frac{1 + 41}{2}

x = \frac{- ( - 1 ) - 41}{2 \cdot 1} : \frac{1 - 41}{2}

The solutions to the quadratic equation are:

x = \frac{1 + 41}{2}, x = \frac{1 - 41}{2}

36 a^{2} - \frac{1}{49} b^{2}

10 x^{2} - 16 x + 5 = 0

w + 50 > 8 w + 78 or \frac{- 89 w + 13}{2} < 5

32 t - 0.8 t^{2} = 160

2 x^{2} - 5 x + 4 = 0

What are the solutions to the equation x^2-x-1=9 ?
The solutions to the equation x^2-x-1=9 are x=(1+sqrt(41))/2 ,x=(1-sqrt(41))/2
Find the zeros of x^2-x-1=9
The zeros of x^2-x-1=9 are x=(1+sqrt(41))/2 ,x=(1-sqrt(41))/2