What are the solutions to the equation 3x^2-3x=11 ?

The solutions to the equation 3x^2-3x=11 are x=(3+sqrt(141))/6 ,x=(3-sqrt(141))/6

Find the zeros of 3x^2-3x=11

The zeros of 3x^2-3x=11 are x=(3+sqrt(141))/6 ,x=(3-sqrt(141))/6

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3 x^{2} - 3 x = 11

x = \frac{3 + 141}{6}, x = \frac{3 - 141}{6}

Decimal Notation

x = 2.47905 \dots, x = - 1.47905 \dots

Solution steps

3 x^{2} - 3 x = 11

Move

11

to the left side

3 x^{2} - 3 x - 11 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 3 ( - 11 )}{2 \cdot 3}

(- 3)^{2} - 4 \cdot 3 (- 11) = 141

x_{1, 2} = \frac{- ( - 3 ) \pm 141}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 141}{2 \cdot 3}, x_{2} = \frac{- ( - 3 ) - 141}{2 \cdot 3}

x = \frac{- ( - 3 ) + 141}{2 \cdot 3} : \frac{3 + 141}{6}

x = \frac{- ( - 3 ) - 141}{2 \cdot 3} : \frac{3 - 141}{6}

The solutions to the quadratic equation are:

x = \frac{3 + 141}{6}, x = \frac{3 - 141}{6}

- x < 3

\frac{\frac{x}{y} - \frac{y}{x}}{\frac{1}{5 x ^{2}} - \frac{1}{5 y ^{2}}}

\frac{73}{18} = \frac{2}{3} n

\frac{x ^{3}}{y ^{2}}

∣ 6 f ∣ = - 2

What are the solutions to the equation 3x^2-3x=11 ?
The solutions to the equation 3x^2-3x=11 are x=(3+sqrt(141))/6 ,x=(3-sqrt(141))/6
Find the zeros of 3x^2-3x=11
The zeros of 3x^2-3x=11 are x=(3+sqrt(141))/6 ,x=(3-sqrt(141))/6