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

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

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

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

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

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

Decimal Notation

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

Solution steps

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

Expand

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

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

Move

2 x^{2}

to the left side

3 x^{2} + 3 x = 2

Move

2

to the left side

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}

9 x^{2} - 6 x + 1 = 7

2 x^{2} = 3 x + 20

x^{2} + 6 x - 1

x, x^{2} y^{2} = 9

(2 x^{2}) 3 = 5832

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