Squares are complete?

We can use quadratic equation to solve this question.

In this quadratic equation: $6x^{2}-31x+35=0$ , we take:

$a=6, b=-31, c=35$

$=\frac{-b+-\sqrt{b^{2}-4ac}}{2a}$

$=\frac{-(-31)+-\sqrt{(-31)^{2}-4\times6\times35}}{2\times6}$

$=\frac{31+-\sqrt{961-840}}{12}$

$=\frac{31+-\sqrt{121}}{12}$

$=\frac{31+-11}{12}$

$=\frac{31+11}{12}$ and $\frac{31-11}{12}$

$=\frac{42}{12}$ and $\frac{20}{12}$

$=\frac{7}{2}$ and $\frac{5}{3}$

So, the roots of the equation are: $\frac{7}{2}$ and $\frac{5}{3}$

Thus, the answer is: $\boxed{\frac{7}{2}, \frac{5}{3}}$

The fastest way to solve an MCQ is to add the two numbers and check if it equals -b/a

$\begin{array}{lr} \text{I solved it with the factorization method :}& \\ 6\varkappa ^{2}-31\varkappa +35 =0& \\ 6\varkappa ^{2}- 10\varkappa -21\varkappa +35 =0 \text{(by breaking the middle term into 2 parts)}& \\ 2\varkappa (3\varkappa -5) -7(3\varkappa -5) =0 \text{(taking a common term)}& \\ (2\varkappa -7)(3\varkappa -5) =0& \\ \text{Now one of the two terms will be equal to zero for the answer to be zero}& \\ 2\varkappa -7 =0 ; 3\varkappa -5 =0& \\ 2\varkappa =7 ; 3\varkappa = 5& \\ \varkappa = \frac{7}{2} ; \varkappa = \frac{5}{3}& \\ \boxed{\frac {7}{2}} ;\boxed {\frac {5}{3}} \end{array}$