What are the solutions to the equation -3x^2-3x+10=0 ?

The solutions to the equation -3x^2-3x+10=0 are x=-(3+sqrt(129))/6 ,x=(sqrt(129)-3)/6

Find the zeros of -3x^2-3x+10=0

The zeros of -3x^2-3x+10=0 are x=-(3+sqrt(129))/6 ,x=(sqrt(129)-3)/6

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- 3 x^{2} - 3 x + 10 = 0

x = - \frac{3 + 129}{6}, x = \frac{129 - 3}{6}

Decimal Notation

x = - 2.39296 \dots, x = 1.39296 \dots

Solution steps

- 3 x^{2} - 3 x + 10 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 ( - 3 ) \cdot 10}{2 ( - 3 )}

(- 3)^{2} - 4 (- 3) \cdot 10 = 129

x_{1, 2} = \frac{- ( - 3 ) \pm 129}{2 ( - 3 )}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 129}{2 ( - 3 )}, x_{2} = \frac{- ( - 3 ) - 129}{2 ( - 3 )}

x = \frac{- ( - 3 ) + 129}{2 ( - 3 )} : - \frac{3 + 129}{6}

x = \frac{- ( - 3 ) - 129}{2 ( - 3 )} : \frac{129 - 3}{6}

The solutions to the quadratic equation are:

x = - \frac{3 + 129}{6}, x = \frac{129 - 3}{6}

6 x^{2} + 10 x + 11 = x^{2}

0 = 96 + 80 t - 16 t^{2}

9 x^{2} - 18 x + 10 = 0

3 x^{2} - 10 x + 9 = 0

(2 x + 3) (x - 2) = 0

What are the solutions to the equation -3x^2-3x+10=0 ?
The solutions to the equation -3x^2-3x+10=0 are x=-(3+sqrt(129))/6 ,x=(sqrt(129)-3)/6
Find the zeros of -3x^2-3x+10=0
The zeros of -3x^2-3x+10=0 are x=-(3+sqrt(129))/6 ,x=(sqrt(129)-3)/6