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sketch each system of equations. Then solve the system by the substitution method
x^2 + y^2=6
y=x^2
~~~~~~~~~~~~~~~~~~
 = 6,     (1)
y = .        (2)
Plots  = 6   and  y = .
To solve the system, replace  in the equation (1) by y, according to (1). You will get
y + y^2 = 6,         (3)   or
 = 0,         (4)
(y+3)*(y-2) = 0  ====>  the roots are  = -3  and   = 2.
Since y = ,  y must be positive, so only the root y = 2 survives.
Then x = +/- .
Answer. The solutions to the system  (1),(2)  are  (x,y) = (,)  and  (x,y) = (,).
To see more solved samples of such systems, look into the lesson
    - Solving the system of algebraic equations of degree 2 
in this site.
Also,  you have this free of charge online textbook in ALGEBRA-I in this site
    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.
The referred lesson is the part of this online textbook under the topic  "Systems of equations that are not linear".