What are the solutions to the equation -3x^2-4x+6=0 ?

The solutions to the equation -3x^2-4x+6=0 are x=-(2+sqrt(22))/3 ,x=(sqrt(22)-2)/3

Find the zeros of -3x^2-4x+6=0

The zeros of -3x^2-4x+6=0 are x=-(2+sqrt(22))/3 ,x=(sqrt(22)-2)/3

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- 3 x^{2} - 4 x + 6 = 0

x = - \frac{2 + 22}{3}, x = \frac{22 - 2}{3}

Decimal Notation

x = - 2.23013 \dots, x = 0.89680 \dots

Solution steps

- 3 x^{2} - 4 x + 6 = 0

Solve with the quadratic formula

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

(- 4)^{2} - 4 (- 3) \cdot 6 = 222

x_{1, 2} = \frac{- ( - 4 ) \pm 2 22}{2 ( - 3 )}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 2 22}{2 ( - 3 )}, x_{2} = \frac{- ( - 4 ) - 2 22}{2 ( - 3 )}

x = \frac{- ( - 4 ) + 2 22}{2 ( - 3 )} : - \frac{2 + 22}{3}

x = \frac{- ( - 4 ) - 2 22}{2 ( - 3 )} : \frac{22 - 2}{3}

The solutions to the quadratic equation are:

x = - \frac{2 + 22}{3}, x = \frac{22 - 2}{3}

3 x^{2} - 25 x + 42 = 0

x, 2 p x^{2} + 6 x + 2 = 0

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

5 x^{2} + 6 x - 4 = 0

x, b^{2} x^{2} - a^{2} y^{2} = a^{2} b^{2}

What are the solutions to the equation -3x^2-4x+6=0 ?
The solutions to the equation -3x^2-4x+6=0 are x=-(2+sqrt(22))/3 ,x=(sqrt(22)-2)/3
Find the zeros of -3x^2-4x+6=0
The zeros of -3x^2-4x+6=0 are x=-(2+sqrt(22))/3 ,x=(sqrt(22)-2)/3