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

The solutions to the equation x=2x^2-2x+3 are x= 3/4+i(sqrt(15))/4 ,x= 3/4-i(sqrt(15))/4

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

The zeros of x=2x^2-2x+3 are x= 3/4+i(sqrt(15))/4 ,x= 3/4-i(sqrt(15))/4

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

x = \frac{3}{4} + i \frac{15}{4}, x = \frac{3}{4} - i \frac{15}{4}

Solution steps

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

Switch sides

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

Move

x

to the left side

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

Solve with the quadratic formula

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

Simplify

(- 3)^{2} - 4 \cdot 2 \cdot 3 : 15 i

x_{1, 2} = \frac{- ( - 3 ) \pm 15 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 15 i}{2 \cdot 2}, x_{2} = \frac{- ( - 3 ) - 15 i}{2 \cdot 2}

x = \frac{- ( - 3 ) + 15 i}{2 \cdot 2} : \frac{3}{4} + i \frac{15}{4}

x = \frac{- ( - 3 ) - 15 i}{2 \cdot 2} : \frac{3}{4} - i \frac{15}{4}

The solutions to the quadratic equation are:

x = \frac{3}{4} + i \frac{15}{4}, x = \frac{3}{4} - i \frac{15}{4}

25 x^{2} - 225 x = 0

12 y^{2} - 9 y - 3 = 0

3 (2 x - 6)^{2} + 6 = 27

12 a^{2} - 22 a - 5 = - 5 a

0 = - x^{2} - 4 x - 4

What are the solutions to the equation x=2x^2-2x+3 ?
The solutions to the equation x=2x^2-2x+3 are x= 3/4+i(sqrt(15))/4 ,x= 3/4-i(sqrt(15))/4
Find the zeros of x=2x^2-2x+3
The zeros of x=2x^2-2x+3 are x= 3/4+i(sqrt(15))/4 ,x= 3/4-i(sqrt(15))/4