8c^2-88c+242 How do you solve this?
\((a-b)^2=a^2-2ab+b^2\)\(8c^2-88c+242=0\ \ \ /:2\\\\4c^2-44c+121=0\\\\(2c)^2-2\cdot2c\cdot11+11^2=0\\\\(2c-11)^2=0\iff2c-11=0\to2c=11\ \ /:2\to c=5.5\)\(8c^2-88c+242=2(4c^2-44c+121)=2\left[(2c)^2-2\cdot2c\cdot11+11^2\right]\\\\=2(2c-11)^2\)