If $\alpha$ and $\beta$ are the roots of the equation $2x^{2}-7x+8=0$, then the equation whose roots are $(3\alpha-4\beta$) and ($3\beta-4\alpha$) is

Given roots alpha, beta of 2x^2 - 7x + 8 = 0, alpha+beta = 3.5, alpha*beta = 4. Let new roots be y1 = 3a-4b, y2 = 3b-4a. Sum = -a-b = -3.5. Product = (3a-4b)(3b-4a) = 9ab - 12a^2 - 12b^2 + 16ab = 25ab - 12(a^2+b^2) = 25ab - 12((a+b)^2 - 2ab) = 25(4) - 12(12.25 - 8) = 100 - 12(4.25) = 100 - 51 = 49. Equation: x^2 - (sum)x + product = x^2 + 3.5x + 49 = 0, or 2x^2 + 7x + 98 = 0.