In the current order of that equation, there is no way to place parentheses/brackets and have it equal 70. I shall explain below.
The factors of 70: 1 and 70, 2 and 35, 5 and 14, 7 and 10
Now, let's experiment with places to put parentheses:
3 x (2 + 4 x 5) = 3(2 + 20) = 3(22) = 66(3 x 2 + 4) x 5 = (6 + 4)(5) = (10)(5) = 503 x (2 + 4) x 5 = 3(6)(5) = (18)(5) = 903 x 2 + (4 x 5) = 6 + 20 = 26(3 x 2) + 4 x 5 = 6 + 20 = 26(3 x 2 + 4 x 5) = (6 + 20) = 26
Notice how the factors of 70 appear nowhere?
The first point is parentheses around the last three constants and it's the closest you'll get. The second point is parentheses around the first three constants and it's up there but not quite 70. The third point is around the middle constants and it's over our number. The fourth, fifth, and sixth points are basically if you were to do the problem straight-up using the order of operations but you added unnecessary parentheses.
The only way this problem would be possible is if you were to switch the places of the 2 and the 4 and make the equation: 3 x 4 + 2 x 5
If it were this, you could do:
(3 x 4 + 2) x 5
= (12 + 2)(5)
= (14)(5)
= 70
The factors 5 and 14 can be made thus satisfying the equality equation.
HOWEVER, I do not believe there are any properties or mathematical principles to allow you to switch the positions of the constants considering we are mixing operations in one equation. If you can think of something that would allow you to switch the positions and maintain the integrity of the equation, that is the only way you can solve this problem.
