What are the solutions to the equation -3x^2-3x+6=0 ?

The solutions to the equation -3x^2-3x+6=0 are x=-2,x=1

Find the zeros of -3x^2-3x+6=0

The zeros of -3x^2-3x+6=0 are x=-2,x=1

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

x = - 2, x = 1

Solution steps

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

Solve with the quadratic formula

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

(- 3)^{2} - 4 (- 3) \cdot 6 = 9

x_{1, 2} = \frac{- ( - 3 ) \pm 9}{2 ( - 3 )}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 9}{2 ( - 3 )}, x_{2} = \frac{- ( - 3 ) - 9}{2 ( - 3 )}

x = \frac{- ( - 3 ) + 9}{2 ( - 3 )} : - 2

x = \frac{- ( - 3 ) - 9}{2 ( - 3 )} : 1

The solutions to the quadratic equation are:

x = - 2, x = 1

2 p \geq 18

36 a^{4} - 108 a^{2} c + 81 c^{2}

\frac{1}{x - 2} \geq \frac{6}{x ^{2} + 2 x - 8}

\frac{\frac{1}{3}}{\frac{2 2}{3}}

7.39 - 4.21 a > 3.65

What are the solutions to the equation -3x^2-3x+6=0 ?
The solutions to the equation -3x^2-3x+6=0 are x=-2,x=1
Find the zeros of -3x^2-3x+6=0
The zeros of -3x^2-3x+6=0 are x=-2,x=1