can anyone help

This is what I did:

1)  I looked at triangle(BAP) and used the Law of Cosines to find angle(BAP).

     I set the length of AB = x.

     BP²  =  AP² + AB² - 2·AP·AB·cos( BAP )

        3²  =  1² + x² - 2·1·x·cos( BAP )

          9  =  1 + x² - 2x·cos( BAP )

         ( 8 - x² ) / ( -2x )  =  cos( BAP )

          cos( BAP ) =  ( x² - 8 ) / (2x )

2)  Then, I looked at triangle(DAP) and used the Law of Cosines to find angle(DAP).

     Since it's a square, AD = x.

     DP²  =  AP² + AD² - 2·AP·AD·cos( DAP )

        7²  =  1² + x² - 2·1·x·cos( DAP )

         49  =  1 + x² - 2x·cos( DAP )

         ( 48 - x² ) / ( -2x )  =  cos( DAP )

          cos( DAP ) =  ( x² - 48 ) / (2x )

3)  Since angle(DAP) = 90^o - angle(BAP)   --->   cos( DAP )  =  sin( BAP )

                      --->   sin( BAP )  =  ( x² - 48 ) / (2x )

4)  Since  sin²( BAP ) + cos²( BAP )  =  1

                [ ( x² - 48 ) / (2x ) ]²  +  [ ( x² - 8 ) / (2x ) ]²  =  1

           [ ( x⁴ - 96x² + 2304 ) / ( 4x² ) ]  +  [ ( x⁴ - 16x² + 64 ) / ( 4x² ) ]  =  1

multiplying both sides by 4x²:

            ( x⁴ - 96x² + 2304 )  +  ( x⁴ - 16x² + 64 )  =  4x²

                 2x⁴ - 116x²  + 2368  =  0

But this doesn't have any real solutions?

Where did I mess up?

ABCD is a rectangle. Given PA = 1; PB=3 and PD=7​, find the area of rectangle ABCD.

Let angle BAP be 45º

Draw the altitudes from point P to point M on side AB, and to point N on side AD.

MP = NP = sqrt( 0.5) = 0.707106781

MB = sqrt ( 3² - 0.5 ) = 2.915475947

ND = sqrt ( 7² - 0.5 ) = 6.964194139

Area  = ( AN + ND )*( AM + MB )