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 = 0.30277 \dots, r = - 3.30277 \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}

2 x y - 2 x^{2} y^{2}

x + 2 = 0

x y (x - 2 y) + 3 y (2 y - x)^{2}

t e^{- 2 t}

\frac{e}{i} = r

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