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

The solutions to the equation 5r^2+r-2=0 are r=(-1+sqrt(41))/(10),r=(-1-sqrt(41))/(10)

Find the zeros of 5r^2+r-2=0

The zeros of 5r^2+r-2=0 are r=(-1+sqrt(41))/(10),r=(-1-sqrt(41))/(10)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

5 r^{2} + r - 2 = 0

r = \frac{- 1 + 41}{10}, r = \frac{- 1 - 41}{10}

Decimal Notation

r = 0.54031 \dots, r = - 0.74031 \dots

Solution steps

5 r^{2} + r - 2 = 0

Solve with the quadratic formula

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

1^{2} - 4 \cdot 5 (- 2) = 41

r_{1, 2} = \frac{- 1 \pm 41}{2 \cdot 5}

Separate the solutions

r_{1} = \frac{- 1 + 41}{2 \cdot 5}, r_{2} = \frac{- 1 - 41}{2 \cdot 5}

r = \frac{- 1 + 41}{2 \cdot 5} : \frac{- 1 + 41}{10}

r = \frac{- 1 - 41}{2 \cdot 5} : \frac{- 1 - 41}{10}

The solutions to the quadratic equation are:

r = \frac{- 1 + 41}{10}, r = \frac{- 1 - 41}{10}

- 7 x^{2} - 10 = - 12 x

x, 7 x^{2} - 5 x y + y^{3} = 3

x^{2} + 10 x + 29

x^{2} - 11 x - 46 = - 6 x + 4

y, x^{2} + x y = ln (y)

What are the solutions to the equation 5r^2+r-2=0 ?
The solutions to the equation 5r^2+r-2=0 are r=(-1+sqrt(41))/(10),r=(-1-sqrt(41))/(10)
Find the zeros of 5r^2+r-2=0
The zeros of 5r^2+r-2=0 are r=(-1+sqrt(41))/(10),r=(-1-sqrt(41))/(10)