What are the solutions to the equation 1/7 x^2+1/21 x-1=0 ?

The solutions to the equation 1/7 x^2+1/21 x-1=0 are x=(-1+sqrt(253))/6 ,x=(-1-sqrt(253))/6

Find the zeros of 1/7 x^2+1/21 x-1=0

The zeros of 1/7 x^2+1/21 x-1=0 are x=(-1+sqrt(253))/6 ,x=(-1-sqrt(253))/6

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

\frac{1}{7} x^{2} + \frac{1}{21} x - 1 = 0

x = \frac{- 1 + 253}{6}, x = \frac{- 1 - 253}{6}

Decimal Notation

x = 2.48432 \dots, x = - 2.81766 \dots

Solution steps

\frac{1}{7} x^{2} + \frac{1}{21} x - 1 = 0

Find Least Common Multiplier of

7, 21 : 21

Multiply by LCM=

21

\frac{1}{7} x^{2} \cdot 21 + \frac{1}{21} x \cdot 21 - 1 \cdot 21 = 0 \cdot 21

Simplify

3 x^{2} + x - 21 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 3 ( - 21 )}{2 \cdot 3}

1^{2} - 4 \cdot 3 (- 21) = 253

x_{1, 2} = \frac{- 1 \pm 253}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 1 + 253}{2 \cdot 3}, x_{2} = \frac{- 1 - 253}{2 \cdot 3}

x = \frac{- 1 + 253}{2 \cdot 3} : \frac{- 1 + 253}{6}

x = \frac{- 1 - 253}{2 \cdot 3} : \frac{- 1 - 253}{6}

The solutions to the quadratic equation are:

x = \frac{- 1 + 253}{6}, x = \frac{- 1 - 253}{6}

\frac{7 π}{6} + 2 π

\frac{3}{4} \cdot \frac{2}{3}

12 m^{2} - 5 m

(\frac{7}{18})^{2}

5 x + 2 = 4 x - 9

What are the solutions to the equation 1/7 x^2+1/21 x-1=0 ?
The solutions to the equation 1/7 x^2+1/21 x-1=0 are x=(-1+sqrt(253))/6 ,x=(-1-sqrt(253))/6
Find the zeros of 1/7 x^2+1/21 x-1=0
The zeros of 1/7 x^2+1/21 x-1=0 are x=(-1+sqrt(253))/6 ,x=(-1-sqrt(253))/6