Question

Solve for r:

$41=400$ $[[\frac{100+r}{100}{]}^2-1]$

Answer 1

$41=400\left[\frac{10000+r^2+200r-100}{100}\right]$

$41=4[r^2+200r+9900]$

$41=4r^2+800r+39600$

$4r^2+800r+39559=0$

$4r^2+442r+358r+39559=0$

$2r(2r+221)+179(2r+221)=0$

$(2r+221)(2r+179)=0$

$\therefore r=\frac{-221}{2},r=\frac{-179}{2}$.