What are the solutions to the equation 3x^2-2x=-5-2x^2 ?

The solutions to the equation 3x^2-2x=-5-2x^2 are x= 1/5+i(2sqrt(6))/5 ,x= 1/5-i(2sqrt(6))/5

Find the zeros of 3x^2-2x=-5-2x^2

The zeros of 3x^2-2x=-5-2x^2 are x= 1/5+i(2sqrt(6))/5 ,x= 1/5-i(2sqrt(6))/5

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3 x^{2} - 2 x = - 5 - 2 x^{2}

x = \frac{1}{5} + i \frac{2 6}{5}, x = \frac{1}{5} - i \frac{2 6}{5}

Solution steps

3 x^{2} - 2 x = - 5 - 2 x^{2}

Move

2 x^{2}

to the left side

5 x^{2} - 2 x = - 5

Move

5

to the left side

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

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 5 \cdot 5}{2 \cdot 5}

Simplify

(- 2)^{2} - 4 \cdot 5 \cdot 5 : 46 i

x_{1, 2} = \frac{- ( - 2 ) \pm 4 6 i}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 4 6 i}{2 \cdot 5}, x_{2} = \frac{- ( - 2 ) - 4 6 i}{2 \cdot 5}

x = \frac{- ( - 2 ) + 4 6 i}{2 \cdot 5} : \frac{1}{5} + i \frac{2 6}{5}

x = \frac{- ( - 2 ) - 4 6 i}{2 \cdot 5} : \frac{1}{5} - i \frac{2 6}{5}

The solutions to the quadratic equation are:

x = \frac{1}{5} + i \frac{2 6}{5}, x = \frac{1}{5} - i \frac{2 6}{5}

19 x + 70 = - x^{2}

x^{2} - 8 x + 16 = 2 x

x^{2} - 8 x + 16 = 75

x (x + 1) = - 8

4 - (2 x + 1)^{2} = (x - 3) - 5 x^{2}

What are the solutions to the equation 3x^2-2x=-5-2x^2 ?
The solutions to the equation 3x^2-2x=-5-2x^2 are x= 1/5+i(2sqrt(6))/5 ,x= 1/5-i(2sqrt(6))/5
Find the zeros of 3x^2-2x=-5-2x^2
The zeros of 3x^2-2x=-5-2x^2 are x= 1/5+i(2sqrt(6))/5 ,x= 1/5-i(2sqrt(6))/5