Question 1207759
{{{pi(1 + x)^2 = 2 + pi(1 + x)}}}


{{{pi(1+2x+x^2)= 2+pi+pi*x}}}


{{{pi +2pi*x+pix^2=2+pi*x+pi}}}

{{{pix^2+pi*x-2=0}}}

Use formula method

a=pi ,b=pi ,c=-2


b^2-4ac = pi^2+8*pi


{{{x =( -pi+-sqrt(pi^2+8pi))/2pi}}}



{{{x =( -pi+-pi *sqrt(1+(8/pi)))/2pi}}}

cancel off the pi


{{{x =( -1+-sqrt(1+(8/pi)))/2}}}

find the value of x