If roots of equation $2{x}^{2}+bx+c=0;b,c\in R$, are real & distinct then the roots of equation $2{cx}^{2}+(b-4c)x+2c-b+1=0$ are

Since the roots of 2x^2+bx+c=0 are real and distinct, its discriminant b^2-8c is greater than zero. The discriminant of the second equation, 2cx^2+(b-4c)x+2c-b+1=0, is (b-4c)^2 - 8c(2c-b+1). Expanding and simplifying this gives b^2-8c, which we know is positive. Because the discriminant of the second equation is positive, its roots are real and distinct.

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