What are the solutions to the equation -x^2-11/3 x+2=0 ?

The solutions to the equation -x^2-11/3 x+2=0 are x=-(11+sqrt(193))/6 ,x=(sqrt(193)-11)/6

Find the zeros of -x^2-11/3 x+2=0

The zeros of -x^2-11/3 x+2=0 are x=-(11+sqrt(193))/6 ,x=(sqrt(193)-11)/6

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- x^{2} - \frac{11}{3} x + 2 = 0

x = - \frac{11 + 193}{6}, x = \frac{193 - 11}{6}

Decimal Notation

x = - 4.14874 \dots, x = 0.48207 \dots

Solution steps

- x^{2} - \frac{11}{3} x + 2 = 0

Multiply both sides by

3

- 3 x^{2} - 11 x + 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 ( - 3 ) \cdot 6}{2 ( - 3 )}

(- 11)^{2} - 4 (- 3) \cdot 6 = 193

x_{1, 2} = \frac{- ( - 11 ) \pm 193}{2 ( - 3 )}

Separate the solutions

x_{1} = \frac{- ( - 11 ) + 193}{2 ( - 3 )}, x_{2} = \frac{- ( - 11 ) - 193}{2 ( - 3 )}

x = \frac{- ( - 11 ) + 193}{2 ( - 3 )} : - \frac{11 + 193}{6}

x = \frac{- ( - 11 ) - 193}{2 ( - 3 )} : \frac{193 - 11}{6}

The solutions to the quadratic equation are:

x = - \frac{11 + 193}{6}, x = \frac{193 - 11}{6}

x, f = 9 x^{2} - 5 x y + 7 y^{2} - 120

k^{2} + 2 k + 3 = 0

1 0^{2} + 1 3^{2} = x^{2}

x, p^{2} x^{2} + (p^{2} - q^{2}) x - q^{2} = 0

25 = 7 + 35 t - 16 t^{2}

What are the solutions to the equation -x^2-11/3 x+2=0 ?
The solutions to the equation -x^2-11/3 x+2=0 are x=-(11+sqrt(193))/6 ,x=(sqrt(193)-11)/6
Find the zeros of -x^2-11/3 x+2=0
The zeros of -x^2-11/3 x+2=0 are x=-(11+sqrt(193))/6 ,x=(sqrt(193)-11)/6