[tex]\displaystyle
(3x-1)+(3x-4)+ (3x-7)+..+(3x-58)=790 \\
\text{Calculam numarul de termeni ai sumei: } \\ \\
n = \frac{58-1}{3}+1 = \frac{57}{3}+1 =19+1 = 20 ~de~termeni \\ \\
(3x-1)+(3x-4)+ (3x-7)+..+(3x-58)=790 \\
3x+3x+3x+ \hdots de ~20~de~ori~ \hdots+3x -1-4-7- \hdots - 58 = 790\\
20\cdot 3x - (1+4+7+ \hdots 58) = 790 \\ \\
20\cdot 3x - \frac{20(58+1)}{2} =790 \\
20\cdot 3x - 10 \cdot 59 =790 \\
60x - 590=790\\
60x=790 +590 \\
60x= 1380 [/tex]
[tex]\displaystyle
x= \frac{1380}{60} = \boxed{23}[/tex]