If x is real, then the maximum and minimum values of the expression x2-3x+4x2+3x+4 will be

Let y=x2-3x+4x2+3x+4⇒(y-1)x2+3(y+1)x+4(y-1)=0For x is real D≥0⇒9(y+1)2-16(y-1)2≥0 ⇒-7y2+50y-7⩾0⇒7y2-50y+7≤0 ⇒(y-7)(7y-1)⩽0Now, the product of two factors is negative if one in -ve and one in +ve.Case I: (y-7)≥0 and (7y-1)≤0 ⇒y≥7 and y≤17But it is impossibleCase II: (y-7)≤0 and (7y-1)≥0⇒y≤7 and y≥17⇒17≤y≤7Hence maximum value is 7 and minimum value is 17