Let $a_{i,j}x^2 + b_{i,j}x + c_{i,j}$ be the trinomial written in cell $(i,j)$ of the table, where $a_{i,j} > 0$ for $1\leq i,j \leq 6$, and \[\{a_{i,j}, b_{i,j}, c_{i,j} \mid 1\leq i,j \leq 6\} = \{-60,-59,\ldots,-1,0,1,\ldots,46,47\}.\] Let $\displaystyle P_j(x) = \alpha_jx^2 + \beta_jx + \gamma_j = \left (\sum_{i=1}^6 a_{i,j} \right )x^2 + \left (\sum_{i=1}^6 b_{i,j} \right )x + \left (\sum_{i=1}^6 c_{i,j} \right )$ be the trinomial which is the sum of the trinomials of column $j$, for $1\leq j \leq 6$, and let $\displaystyle P(x) = \alpha x^2 + \beta x + \gamma = \sum_{j=1}^6 P_j(x)$. Suppose none of the trinomials $P_j(x)$ has real roots; then $P_j(x) > 0$ for all $1\leq j \leq 6$ and $x\in \mathbb{R}$ (since their leading coefficients are positive). But then a fortiori $P(x) > 0$ for all $x\in \mathbb{R}$, which is equivalent to $\Delta = \beta^2 - 4\alpha \gamma < 0$. But $\displaystyle \alpha + \beta + \gamma = - \sum_{k=1}^{60} k + \sum_{k=1}^{47} k = -702 = -v$. Then \[\Delta = \beta^2 - 4\alpha \gamma = (v+\alpha+\gamma)^2 - 4\alpha \gamma = (v - \alpha + \gamma)^2 + 4v\alpha > 0,\] since $v>0$ and $\alpha > 0$. We arrived at a clear contradiction. Remark. The only thing that mattered is therefore that the negative bound ($-60$) for the values of the coefficients was given larger in absolute value than the positive one ($47$), leading to $v>0$. Another problem that looks a bit strange.