Solve `(x^2-5x+7)-(x-2)(x-3)=1.`

Explanation

We have, `(x^(2) -5x+7)^(2)- (x-2)(x-3) =1`
or `(x^(2) -5x+7)^(2)- (x^(2)-5x+7)=0`
`rArr Y^(2) - Y = 0`, Where `y = x^(2) - 5x + 7`
or y(y-1)=0
Now, y = 0
`rArr x^(2) - 5x + 7 = 0`
or `x = 5pmsqrt(25-28)/(2) = (5pmsqrt-3)/(2)=(5pmisqrt3)/(2)`
y = 1 (where i = `sqrt(-1)` and y = 1)`
`rArr x^(2) - 5x + 6= 0`
or (x-3) (x-2) = 0
or x = 3, 2
Hence, the roots of the equation are 2, 3, `(5+isqrt(3))//2, and (5-isqrt(3))//2`.