You could skip a step by using SolveValues:

{x1, x2} = SolveValues[x^2 + a x + 1 == 0, x]

Further question: I want to assign a root greater than 0 to y1, and write the following code. Why is the result not what you want, and the calculated value of y1 less than 0? Why? How to rewrite the code?

ClearAll[Evaluate[Context[] <> "*"]]

eqns1 = {x^2/2^2 + y^2/1^2 == 1, y == x + 1};  

{{x1, y1}, {x2, y2}} = 
 SolveValues[eqns1, {x, y}, Assumptions -> y1 > 0] // FullSimplify

the result is

{{-(8/5), -(3/5)}, {0, 1}}

I want to write all solution sets, but assign the value of y greater than 0 to y1, and other solutions will also be displayed.

eqns1 = {x^2/2^2 + y^2/1^2 == 1, y == x + 1};

{{x1, y1}} = SolveValues[eqns1, {x, y}, Assumptions -> y > 0]

{{x2, y2}} = SolveValues[eqns1, {x, y}, Assumptions -> y < 0]

This can solve the problem, but the code is awkward. Can you optimize it?

ClearAll[Evaluate[Context[] <> "*"]]

eqns1 = {x^2/2^2 + y^2/1^2 == 1, y == x + 1};   

SolveValues[eqns1, {x, y}]

{{x1, y1}, {x2, y2}} = Sort[%, #1[[2]] < #2[[2]] &]

Or

sol = SolveValues[eqns1, {x, y}]
{{x1, y1}, {x2, y2}} = SortBy[Last]

Oops. I meant to write

{{x1, y1}, {x2, y2}} = sol // SortBy[Last]