What are the solutions to the equation r^2-3r-1=0 ?

The solutions to the equation r^2-3r-1=0 are r=(3+sqrt(13))/2 ,r=(3-sqrt(13))/2

Find the zeros of r^2-3r-1=0

The zeros of r^2-3r-1=0 are r=(3+sqrt(13))/2 ,r=(3-sqrt(13))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

r^{2} - 3 r - 1 = 0

r = \frac{3 + 13}{2}, r = \frac{3 - 13}{2}

Decimal Notation

r = 3.30277 \dots, r = - 0.30277 \dots

Solution steps

r^{2} - 3 r - 1 = 0

Solve with the quadratic formula

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

(- 3)^{2} - 4 \cdot 1 \cdot (- 1) = 13

r_{1, 2} = \frac{- ( - 3 ) \pm 13}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- ( - 3 ) + 13}{2 \cdot 1}, r_{2} = \frac{- ( - 3 ) - 13}{2 \cdot 1}

r = \frac{- ( - 3 ) + 13}{2 \cdot 1} : \frac{3 + 13}{2}

r = \frac{- ( - 3 ) - 13}{2 \cdot 1} : \frac{3 - 13}{2}

The solutions to the quadratic equation are:

r = \frac{3 + 13}{2}, r = \frac{3 - 13}{2}

3 x^{2} + 32 x + 80 = 0

q^{2} - 4 = 0

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

x, z = y^{2} - 3 x y + 2 x^{2} - 4

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

What are the solutions to the equation r^2-3r-1=0 ?
The solutions to the equation r^2-3r-1=0 are r=(3+sqrt(13))/2 ,r=(3-sqrt(13))/2
Find the zeros of r^2-3r-1=0
The zeros of r^2-3r-1=0 are r=(3+sqrt(13))/2 ,r=(3-sqrt(13))/2