What are the solutions to the equation 5x^2-3(2x-1)=2x^2-8x ?

The solutions to the equation 5x^2-3(2x-1)=2x^2-8x are x=-1/3+i(2sqrt(2))/3 ,x=-1/3-i(2sqrt(2))/3

Find the zeros of 5x^2-3(2x-1)=2x^2-8x

The zeros of 5x^2-3(2x-1)=2x^2-8x are x=-1/3+i(2sqrt(2))/3 ,x=-1/3-i(2sqrt(2))/3

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5 x^{2} - 3 (2 x - 1) = 2 x^{2} - 8 x

x = - \frac{1}{3} + i \frac{2 2}{3}, x = - \frac{1}{3} - i \frac{2 2}{3}

Solution steps

5 x^{2} - 3 (2 x - 1) = 2 x^{2} - 8 x

Expand

5 x^{2} - 3 (2 x - 1) : 5 x^{2} - 6 x + 3

5 x^{2} - 6 x + 3 = 2 x^{2} - 8 x

Move

8 x

to the left side

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

Move

2 x^{2}

to the left side

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

Solve with the quadratic formula

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

Simplify

2^{2} - 4 \cdot 3 \cdot 3 : 42 i

x_{1, 2} = \frac{- 2 \pm 4 2 i}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 2 + 4 2 i}{2 \cdot 3}, x_{2} = \frac{- 2 - 4 2 i}{2 \cdot 3}

x = \frac{- 2 + 4 2 i}{2 \cdot 3} : - \frac{1}{3} + i \frac{2 2}{3}

x = \frac{- 2 - 4 2 i}{2 \cdot 3} : - \frac{1}{3} - i \frac{2 2}{3}

The solutions to the quadratic equation are:

x = - \frac{1}{3} + i \frac{2 2}{3}, x = - \frac{1}{3} - i \frac{2 2}{3}

8 x^{2} - 24 = 8

2 x^{2} - 3 x + 4 = x^{2} + 5 x - 3

3 x + x^{2} - 33 - x 3 = 0

x, x^{2} + 5 x + 2 = - 4 es

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

What are the solutions to the equation 5x^2-3(2x-1)=2x^2-8x ?
The solutions to the equation 5x^2-3(2x-1)=2x^2-8x are x=-1/3+i(2sqrt(2))/3 ,x=-1/3-i(2sqrt(2))/3
Find the zeros of 5x^2-3(2x-1)=2x^2-8x
The zeros of 5x^2-3(2x-1)=2x^2-8x are x=-1/3+i(2sqrt(2))/3 ,x=-1/3-i(2sqrt(2))/3