What are the solutions to the equation 11x=3x^2-9 ?

The solutions to the equation 11x=3x^2-9 are x=(11+sqrt(229))/6 ,x=(11-sqrt(229))/6

Find the zeros of 11x=3x^2-9

The zeros of 11x=3x^2-9 are x=(11+sqrt(229))/6 ,x=(11-sqrt(229))/6

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11 x = 3 x^{2} - 9

x = \frac{11 + 229}{6}, x = \frac{11 - 229}{6}

Decimal Notation

x = 4.35545 \dots, x = - 0.68879 \dots

Solution steps

11 x = 3 x^{2} - 9

Switch sides

3 x^{2} - 9 = 11 x

Move

11 x

to the left side

3 x^{2} - 9 - 11 x = 0

Write in the standard form

a x^{2} + b x + c = 0

3 x^{2} - 11 x - 9 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 \cdot 3 ( - 9 )}{2 \cdot 3}

(- 11)^{2} - 4 \cdot 3 (- 9) = 229

x_{1, 2} = \frac{- ( - 11 ) \pm 229}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 11 ) + 229}{2 \cdot 3}, x_{2} = \frac{- ( - 11 ) - 229}{2 \cdot 3}

x = \frac{- ( - 11 ) + 229}{2 \cdot 3} : \frac{11 + 229}{6}

x = \frac{- ( - 11 ) - 229}{2 \cdot 3} : \frac{11 - 229}{6}

The solutions to the quadratic equation are:

x = \frac{11 + 229}{6}, x = \frac{11 - 229}{6}

x, x^{2} - 4 x + 2^{t} = 0

x, x^{2} + y^{2} + 2 x = - 1

2 d^{2} = 8 d + 90

2 x^{2} + 3 x - 16 = 4

v^{2} = 47

What are the solutions to the equation 11x=3x^2-9 ?
The solutions to the equation 11x=3x^2-9 are x=(11+sqrt(229))/6 ,x=(11-sqrt(229))/6
Find the zeros of 11x=3x^2-9
The zeros of 11x=3x^2-9 are x=(11+sqrt(229))/6 ,x=(11-sqrt(229))/6