Solve for \(f\) in \(f\times 4=7\).
Solve for \(f\).
\[f\times 4=7\]
Regroup terms.
\[4f=7\]
Divide both sides by \(4\).
\[f=\frac{7}{4}\]
\[f=\frac{7}{4}\]
Substitute \(f=\frac{7}{4}\) into \(fx=5{x}^{2}+4\).
Start with the original equation.
\[fx=5{x}^{2}+4\]
Let \(f=\frac{7}{4}\).
\[\frac{7}{4}x=5{x}^{2}+4\]
Simplify.
\[\frac{7x}{4}=5{x}^{2}+4\]
\[\frac{7x}{4}=5{x}^{2}+4\]
Solve for \(x\) in \(\frac{7x}{4}=5{x}^{2}+4\).
Solve for \(x\).
\[\frac{7x}{4}=5{x}^{2}+4\]
Multiply both sides by \(4\).
\[7x=20{x}^{2}+16\]
Move all terms to one side.
\[7x-20{x}^{2}-16=0\]
Use the Quadratic Formula.
In general, given \(a{x}^{2}+bx+c=0\), there exists two solutions where:
\[x=\frac{-b+\sqrt{{b}^{2}-4ac}}{2a},\frac{-b-\sqrt{{b}^{2}-4ac}}{2a}\]
In this case, \(a=-20\), \(b=7\) and \(c=-16\).
\[{x}^{}=\frac{-7+\sqrt{{7}^{2}-4\times -20\times -16}}{2\times -20},\frac{-7-\sqrt{{7}^{2}-4\times -20\times -16}}{2\times -20}\]
Simplify.
\[x=\frac{-7+\sqrt{1231}\imath }{-40},\frac{-7-\sqrt{1231}\imath }{-40}\]
\[x=\frac{-7+\sqrt{1231}\imath }{-40},\frac{-7-\sqrt{1231}\imath }{-40}\]
Simplify solutions.
\[x=-\frac{-7+\sqrt{1231}\imath }{40},-\frac{-7-\sqrt{1231}\imath }{40}\]
\[x=-\frac{-7+\sqrt{1231}\imath }{40},-\frac{-7-\sqrt{1231}\imath }{40}\]
Therefore,
\[\begin{aligned}&f=\frac{7}{4}\\&x=-\frac{-7+\sqrt{1231}\imath }{40},-\frac{-7-\sqrt{1231}\imath }{40}\end{aligned}\]
f=7/4;x=-(-7+sqrt(1231)*IM)/40,-(-7-sqrt(1231)*IM)/40
