What are the solutions to the equation 3x^2-(-4x+7)=2x^2+5x-9 ?

The solutions to the equation 3x^2-(-4x+7)=2x^2+5x-9 are x= 1/2+i(sqrt(7))/2 ,x= 1/2-i(sqrt(7))/2

Find the zeros of 3x^2-(-4x+7)=2x^2+5x-9

The zeros of 3x^2-(-4x+7)=2x^2+5x-9 are x= 1/2+i(sqrt(7))/2 ,x= 1/2-i(sqrt(7))/2

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3 x^{2} - (- 4 x + 7) = 2 x^{2} + 5 x - 9

x = \frac{1}{2} + i \frac{7}{2}, x = \frac{1}{2} - i \frac{7}{2}

Solution steps

3 x^{2} - (- 4 x + 7) = 2 x^{2} + 5 x - 9

Expand

3 x^{2} - (- 4 x + 7) : 3 x^{2} + 4 x - 7

3 x^{2} + 4 x - 7 = 2 x^{2} + 5 x - 9

Move

9

to the left side

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

Move

5 x

to the left side

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

Move

2 x^{2}

to the left side

x^{2} - x + 2 = 0

Solve with the quadratic formula

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

Simplify

(- 1)^{2} - 4 \cdot 1 \cdot 2 : 7 i

x_{1, 2} = \frac{- ( - 1 ) \pm 7 i}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 7 i}{2 \cdot 1}, x_{2} = \frac{- ( - 1 ) - 7 i}{2 \cdot 1}

x = \frac{- ( - 1 ) + 7 i}{2 \cdot 1} : \frac{1}{2} + i \frac{7}{2}

x = \frac{- ( - 1 ) - 7 i}{2 \cdot 1} : \frac{1}{2} - i \frac{7}{2}

The solutions to the quadratic equation are:

x = \frac{1}{2} + i \frac{7}{2}, x = \frac{1}{2} - i \frac{7}{2}

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

3 w^{2} - w + 5 = 6

x^{2} - 3 x - 1 = 0

x^{2} = 3

4 (6 a - 3)^{2} + 28 = 32

What are the solutions to the equation 3x^2-(-4x+7)=2x^2+5x-9 ?
The solutions to the equation 3x^2-(-4x+7)=2x^2+5x-9 are x= 1/2+i(sqrt(7))/2 ,x= 1/2-i(sqrt(7))/2
Find the zeros of 3x^2-(-4x+7)=2x^2+5x-9
The zeros of 3x^2-(-4x+7)=2x^2+5x-9 are x= 1/2+i(sqrt(7))/2 ,x= 1/2-i(sqrt(7))/2