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

The solutions to the equation -x+3=-2x^2 are x= 1/4-i(sqrt(23))/4 ,x= 1/4+i(sqrt(23))/4

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

The zeros of -x+3=-2x^2 are x= 1/4-i(sqrt(23))/4 ,x= 1/4+i(sqrt(23))/4

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

x = \frac{1}{4} - i \frac{23}{4}, x = \frac{1}{4} + i \frac{23}{4}

Solution steps

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

Switch sides

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

Move

3

to the left side

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

Move

x

to the left side

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

Write in the standard form

a x^{2} + b x + c = 0

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

Solve with the quadratic formula

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

Simplify

1^{2} - 4 (- 2) (- 3) : 23 i

x_{1, 2} = \frac{- 1 \pm 23 i}{2 ( - 2 )}

Separate the solutions

x_{1} = \frac{- 1 + 23 i}{2 ( - 2 )}, x_{2} = \frac{- 1 - 23 i}{2 ( - 2 )}

x = \frac{- 1 + 23 i}{2 ( - 2 )} : \frac{1}{4} - i \frac{23}{4}

x = \frac{- 1 - 23 i}{2 ( - 2 )} : \frac{1}{4} + i \frac{23}{4}

The solutions to the quadratic equation are:

x = \frac{1}{4} - i \frac{23}{4}, x = \frac{1}{4} + i \frac{23}{4}

y^{2} - 18 y + 17

x^{3} - 1331

27 d^{6} + 8 g^{12}

2 (x - 5)^{2} = 90

36 + 16

What are the solutions to the equation -x+3=-2x^2 ?
The solutions to the equation -x+3=-2x^2 are x= 1/4-i(sqrt(23))/4 ,x= 1/4+i(sqrt(23))/4
Find the zeros of -x+3=-2x^2
The zeros of -x+3=-2x^2 are x= 1/4-i(sqrt(23))/4 ,x= 1/4+i(sqrt(23))/4