What are the solutions to the equation 6sqrt(3)*x^2+7x=sqrt(3) ?

The solutions to the equation 6sqrt(3)*x^2+7x=sqrt(3) are x= 1/(3sqrt(3)),x=-(sqrt(3))/2

Find the zeros of 6sqrt(3)*x^2+7x=sqrt(3)

The zeros of 6sqrt(3)*x^2+7x=sqrt(3) are x= 1/(3sqrt(3)),x=-(sqrt(3))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

63 \cdot x^{2} + 7 x = 3

x = \frac{1}{3 3}, x = - \frac{3}{2}

Decimal Notation

x = 0.19245 \dots, x = - 0.86602 \dots

Solution steps

63 x^{2} + 7 x = 3

Move

3

to the left side

63 x^{2} + 7 x - 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 6 3 ( - 3 )}{2 \cdot 6 3}

7^{2} - 4 \cdot 63 (- 3) = 11

x_{1, 2} = \frac{- 7 \pm 11}{2 \cdot 6 3}

Separate the solutions

x_{1} = \frac{- 7 + 11}{2 \cdot 6 3}, x_{2} = \frac{- 7 - 11}{2 \cdot 6 3}

x = \frac{- 7 + 11}{2 \cdot 6 3} : \frac{1}{3 3}

x = \frac{- 7 - 11}{2 \cdot 6 3} : - \frac{3}{2}

The solutions to the quadratic equation are:

x = \frac{1}{3 3}, x = - \frac{3}{2}

n, p = (n^{2}) + 1

(\frac{x + 1}{3}) (\frac{x - 1}{3}) = \frac{1}{3}

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

x^{3} - 4 x^{2} - x + 1 = (x - 2)^{3}

x, y^{2} - 1 = (x - 1) (x - 3)

What are the solutions to the equation 6sqrt(3)*x^2+7x=sqrt(3) ?
The solutions to the equation 6sqrt(3)*x^2+7x=sqrt(3) are x= 1/(3sqrt(3)),x=-(sqrt(3))/2
Find the zeros of 6sqrt(3)*x^2+7x=sqrt(3)
The zeros of 6sqrt(3)*x^2+7x=sqrt(3) are x= 1/(3sqrt(3)),x=-(sqrt(3))/2