What are the solutions to the equation 3x^2+(2+1)x-2=0 ?

The solutions to the equation 3x^2+(2+1)x-2=0 are x=(-3+sqrt(33))/6 ,x=(-3-sqrt(33))/6

Find the zeros of 3x^2+(2+1)x-2=0

The zeros of 3x^2+(2+1)x-2=0 are x=(-3+sqrt(33))/6 ,x=(-3-sqrt(33))/6

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3 x^{2} + (2 + 1) x - 2 = 0

x = \frac{- 3 + 33}{6}, x = \frac{- 3 - 33}{6}

Decimal Notation

x = 0.45742 \dots, x = - 1.45742 \dots

Solution steps

3 x^{2} + (2 + 1) x - 2 = 0

Expand

3 x^{2} + (2 + 1) x - 2 : 3 x^{2} + 3 x - 2

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

Solve with the quadratic formula

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

3^{2} - 4 \cdot 3 (- 2) = 33

x_{1, 2} = \frac{- 3 \pm 33}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 3 + 33}{2 \cdot 3}, x_{2} = \frac{- 3 - 33}{2 \cdot 3}

x = \frac{- 3 + 33}{2 \cdot 3} : \frac{- 3 + 33}{6}

x = \frac{- 3 - 33}{2 \cdot 3} : \frac{- 3 - 33}{6}

The solutions to the quadratic equation are:

x = \frac{- 3 + 33}{6}, x = \frac{- 3 - 33}{6}

\frac{x ^{2}}{3} - x = 7

x, 3 f (1 - x) - f = 3 x^{2} - 1

x^{2} + 5 = - 6 x

r, C = 2 π r^{2}

\frac{9}{25} - x^{2} = 0

What are the solutions to the equation 3x^2+(2+1)x-2=0 ?
The solutions to the equation 3x^2+(2+1)x-2=0 are x=(-3+sqrt(33))/6 ,x=(-3-sqrt(33))/6
Find the zeros of 3x^2+(2+1)x-2=0
The zeros of 3x^2+(2+1)x-2=0 are x=(-3+sqrt(33))/6 ,x=(-3-sqrt(33))/6