The function will be continuous if the piecewise components are equal at their boundaries.

Thus you want to solve the following system:

$$a (0)+2 b = (0)^2 +3 a - b$$ $$ (2)^2+3a-b = 3 (2) -5 $$

You could input this as Solve[{a0+2 b = 0^2 +3 a - b, 2^2+3a-b = 32 -5},{a,b}]

Though this shouldn't require Wolfram's help, as you can simplify the system pretty easily!

$$a (0)+2 b = 2b$$ $$(0)^2 +3 a - b= 3a-b$$ $$ 2b= 3a-b \Longrightarrow b=a $$ $$ (2)^2+3a-b = 4 +3 a-b= 4+2 a$$ $$3 (2) -5 = 6-5 = 1$$ $$4+2a = 1 \Longrightarrow 2a = -3 \Longrightarrow a =-3/2 =b$$