What are the solutions to the equation (x-2)(3-2x)=3 ?

The solutions to the equation (x-2)(3-2x)=3 are x= 7/4-i(sqrt(23))/4 ,x= 7/4+i(sqrt(23))/4

Find the zeros of (x-2)(3-2x)=3

The zeros of (x-2)(3-2x)=3 are x= 7/4-i(sqrt(23))/4 ,x= 7/4+i(sqrt(23))/4

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(x - 2) (3 - 2 x) = 3

x = \frac{7}{4} - i \frac{23}{4}, x = \frac{7}{4} + i \frac{23}{4}

Solution steps

(x - 2) (3 - 2 x) = 3

Expand

(x - 2) (3 - 2 x) : - 2 x^{2} + 7 x - 6

- 2 x^{2} + 7 x - 6 = 3

Move

3

to the left side

- 2 x^{2} + 7 x - 9 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 ( - 2 ) ( - 9 )}{2 ( - 2 )}

Simplify

7^{2} - 4 (- 2) (- 9) : 23 i

x_{1, 2} = \frac{- 7 \pm 23 i}{2 ( - 2 )}

Separate the solutions

x_{1} = \frac{- 7 + 23 i}{2 ( - 2 )}, x_{2} = \frac{- 7 - 23 i}{2 ( - 2 )}

x = \frac{- 7 + 23 i}{2 ( - 2 )} : \frac{7}{4} - i \frac{23}{4}

x = \frac{- 7 - 23 i}{2 ( - 2 )} : \frac{7}{4} + i \frac{23}{4}

The solutions to the quadratic equation are:

x = \frac{7}{4} - i \frac{23}{4}, x = \frac{7}{4} + i \frac{23}{4}

(2 y + 9)^{2} = 81

x^{2} - 4 x + 4 = 12

A = 4 π x^{2}

b^{2} + g = \frac{A}{m}

(8 b - 7)^{2} = 169

What are the solutions to the equation (x-2)(3-2x)=3 ?
The solutions to the equation (x-2)(3-2x)=3 are x= 7/4-i(sqrt(23))/4 ,x= 7/4+i(sqrt(23))/4
Find the zeros of (x-2)(3-2x)=3
The zeros of (x-2)(3-2x)=3 are x= 7/4-i(sqrt(23))/4 ,x= 7/4+i(sqrt(23))/4