Find p and q, if Definition area is [3;10]

[tex]\\ \tt\longmapsto y=\sqrt{-x^2+px+q}[/tex]
[tex]\\ \tt\longmapsto y^2=-x^2+px+q[/tex]
[tex]\\ \tt\longmapsto 10^2=-(3^)2+3p+q[/tex]
[tex]\\ \tt\longmapsto 100=-9+3p+q[/tex]
[tex]\\ \tt\longmapsto 3p+q=109[/tex]
Let p,q be (27,28)
Apply
[tex]\\ \tt\longmapsto 3(27)+28=81+28=109[/tex]
Verified