What are the solutions to the equation 5x^2+x=1 ?

The solutions to the equation 5x^2+x=1 are x=(-1+sqrt(21))/(10),x=(-1-sqrt(21))/(10)

Find the zeros of 5x^2+x=1

The zeros of 5x^2+x=1 are x=(-1+sqrt(21))/(10),x=(-1-sqrt(21))/(10)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

5 x^{2} + x = 1

x = \frac{- 1 + 21}{10}, x = \frac{- 1 - 21}{10}

Decimal Notation

x = 0.35825 \dots, x = - 0.55825 \dots

Solution steps

5 x^{2} + x = 1

Move

1

to the left side

5 x^{2} + x - 1 = 0

Solve with the quadratic formula

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

1^{2} - 4 \cdot 5 (- 1) = 21

x_{1, 2} = \frac{- 1 \pm 21}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- 1 + 21}{2 \cdot 5}, x_{2} = \frac{- 1 - 21}{2 \cdot 5}

x = \frac{- 1 + 21}{2 \cdot 5} : \frac{- 1 + 21}{10}

x = \frac{- 1 - 21}{2 \cdot 5} : \frac{- 1 - 21}{10}

The solutions to the quadratic equation are:

x = \frac{- 1 + 21}{10}, x = \frac{- 1 - 21}{10}

\frac{3}{7} \cdot 35

2 - 2

(\frac{6 x ^{3}}{3 x ^{5}})^{- 2}

3 - \frac{3}{4} x = \frac{7}{10}

e^{y} \cdot e^{y}

What are the solutions to the equation 5x^2+x=1 ?
The solutions to the equation 5x^2+x=1 are x=(-1+sqrt(21))/(10),x=(-1-sqrt(21))/(10)
Find the zeros of 5x^2+x=1
The zeros of 5x^2+x=1 are x=(-1+sqrt(21))/(10),x=(-1-sqrt(21))/(10)