$\text{( a= 1, b= -4 , c = m-1 )}$

Ta có: Δ$\text{= $b^{2}$ - 4ac}$= $\text{$(-4^{2}$) - 4( m - 1) = 16 -4m + 4 = 20 - 4m}$ 

Để phương trình có $\text{2}$ nghiệm thì: 

$\text{Δ ≥ 0 }$ hay $\text{20 - 4m}$ $\text{≥ 0}$

$\text{-4m ≥ - 20}$

$\text{m ≤ 5 }$

Theo định lý Viete : $\left \{ {{x_{1} + x_{2} = \frac{-b}{a}= 4} \atop {x_{1}x_{2}}= \frac{c}{a}=  m-1} \right.$

Theo đề bài ta có: $x_{1}^{2}$ $\text{+ }$ $x_{2}^{2}$ - $\text{10}$$x_{1}$$x_{2}$ $\text{= 2020}$ 

$\text{(}$$x_{1}$ $\text{+}$$x_{2}$ $\text{)}$$^{2}$ - $\text{10}$$x_{1}$$x_{2}$ - $\text{2}$$x_{1}$$x_{2}$ $\text{= 2020}$ 

$\text{(}$$x_{1}$ $\text{+}$$x_{2}$ $\text{)}$$^{2}$ - $\text{12}$$x_{1}$$x_{2}$$\text{= 2020}$ 

$\text{$4^{2}$ - 12 (m - 1)}$$\text{= 2020}$ 

$\text{16 - 12m + 12 - 2020 = 0}$

$\text{-12m -1992 = 0}$

$\text{-12m = 1992}$

$\text{m= - 166 }$ $\text{( Thỏa mãn)}$

Vậy $\text{m = - 166 }$