which point satisfies the linear quadratic system y=x+3 and y=5-x^2 ?

Option A

On solving both,
x+3=5−x2
⇒x2+x−2=0
⇒x2+2x−x−2=0
⇒x(x+2)x−1(x+2)=0
⇒(x+2)(x−1)=0
⇒x=−2,1
When, x=−2,y=−2+3=1, point is (-2,1)
When, x=1,y=1+3=4, point is (1,4)