\(\dfrac{x\left(x-1\right)}{2}=28\\ \Rightarrow x^2-x-56=0\\ \Rightarrow\left(x^2-x-\dfrac{1}{4}\right)-\dfrac{225}{4}=0\\ \Rightarrow\left(x-\dfrac{1}{2}\right)^2-\dfrac{225}{4}=0\\ \Rightarrow\left(x-8\right)\left(x+7\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=8\\x=-7\end{matrix}\right.\)
Theo đề ta có:
\(\dfrac{x.\left(x-1\right)}{2}=28\)
=> \(x.\left(x-1\right)=28.2\)
=> \(x.\left(x-1\right)=56\)
=> \(x.x-x.1=56\)
=> \(x^2-x\)= 56
=> \(8^2-8\) = 56
Vậy x= 8 thì thỏa mãn đề bài yêu cầu