The answer is a. 3 and 8.
Given the equation
[tex]\displaystyle \frac{6}{x}+\frac{x-3}{4}=2[/tex]
Multiply both sides by [tex]4x[/tex]
[tex]\displaystyle 4x\bigg(\frac{6}{x}+\frac{x-3}{4}=2\bigg)[/tex]
[tex]\implies 4(6)+x(x-3)=4(2x)[/tex]
[tex]\implies 24+x^2-3x=8x[/tex]
With some simplification and rearranging, we have
[tex]x^2-11x+24=0[/tex]
The factors of 24 that has a sum of -11 is -8 and -3, so, it can be factored as
[tex](x-8)(x-3)=0[/tex]
Which implies that
[tex]\begin{array}{c|c} x-8=0 & x-3=0\\ \boxed{x=8}&\boxed{x=3} \end{array}[/tex]
Thus, the roots are 3 and 8.
