Factor with quadratic formula.
In general, given \(a{x}^{2}+bx+c\), the factored form is:
\[a(x-\frac{-b+\sqrt{{b}^{2}-4ac}}{2a})(x-\frac{-b-\sqrt{{b}^{2}-4ac}}{2a})\]
In this case, \(a=49\), \(b=7\) and \(c=\frac{1}{4}\).
\[49(x-\frac{-7+\sqrt{{7}^{2}-4\times 49\times \frac{1}{4}}}{2\times 49})(x-\frac{-7-\sqrt{{7}^{2}-4\times 49\times \frac{1}{4}}}{2\times 49})\]
Simplify.
\[49(x+\frac{1}{14})(x+\frac{1}{14})\]
Solve for \(x\).
Ask: When will \((x+\frac{1}{14})(x+\frac{1}{14})\) equal zero?
When \(x+\frac{1}{14}=0\) or \(x+\frac{1}{14}=0\)
Solve each of the 2 equations above.
\[x=-\frac{1}{14}\]
