\(7x^2-31x+20=0\)
First, remove the coefficient by division
So
\(\frac{7x^2}{7}-\frac{31x}{7}+\frac{20}{7}=0\)
Then, cancel out \(\frac{20}{7}\)
So,
\(x^2-\frac{31x}{7}+\frac{20}{7}-\frac{20}{7}=-\frac{20}{7}\)
Then take half of the coefficient, then square it.
\(\frac{-\frac{31}{7}}{2}=\frac{37}{14}\)
\((-\frac{37}{14})^2=\frac{961}{196}\)
So, The equation already looks like this:
\(x^2-\frac{31}{7}+\frac{961}{196}=-\frac{20}{7}+\frac{961}{196}\)
So,
\(x^2-\frac{31}{7}+\frac{961}{196}=\frac{401}{196}\)
Then factor the left side of the equation:
\((x-\frac{31}{14})^2=\frac{401}{196}\)
Then square both sides:
\(x-\frac{31}{14}=\sqrt{\frac{401}{196}}\)
