What are the solutions to the equation 2x^2+12x=4-x^2 ?

The solutions to the equation 2x^2+12x=4-x^2 are x=(2(2sqrt(3)-3))/3 ,x=-(2(3+2sqrt(3)))/3

Find the zeros of 2x^2+12x=4-x^2

The zeros of 2x^2+12x=4-x^2 are x=(2(2sqrt(3)-3))/3 ,x=-(2(3+2sqrt(3)))/3

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2 x^{2} + 12 x = 4 - x^{2}

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

Solution steps

2 x^{2} + 12 x = 4 - x^{2}

Move

x^{2}

to the left side

3 x^{2} + 12 x = 4

Move

4

to the left side

3 x^{2} + 12 x - 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 3 ( - 4 )}{2 \cdot 3}

1 2^{2} - 4 \cdot 3 (- 4) = 83

x_{1, 2} = \frac{- 12 \pm 8 3}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 12 + 8 3}{2 \cdot 3}, x_{2} = \frac{- 12 - 8 3}{2 \cdot 3}

x = \frac{- 12 + 8 3}{2 \cdot 3} : \frac{2 ( 2 3 - 3 )}{3}

x = \frac{- 12 - 8 3}{2 \cdot 3} : - \frac{2 ( 3 + 2 3 )}{3}

The solutions to the quadratic equation are:

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

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

4 m^{2} - 3 m^{6} + 5 m^{4}

(\frac{1}{16})^{x} - (\frac{1}{4})^{x} \geq 0

(\frac{2}{9})^{0}

- 2 (2 x - 5) = 2 x

What are the solutions to the equation 2x^2+12x=4-x^2 ?
The solutions to the equation 2x^2+12x=4-x^2 are x=(2(2sqrt(3)-3))/3 ,x=-(2(3+2sqrt(3)))/3
Find the zeros of 2x^2+12x=4-x^2
The zeros of 2x^2+12x=4-x^2 are x=(2(2sqrt(3)-3))/3 ,x=-(2(3+2sqrt(3)))/3