Answer
The solution of the equation $3{{x}^{2}}-7x+1=0$ is $x=\frac{7+\sqrt{37}}{6}$ or $x=\frac{7-\sqrt{37}}{6}$.
Work Step by Step
Consider the provided equation
$3{{x}^{2}}-7x+1=0$
Compare the equation $3{{x}^{2}}-7x+1=0$ with the standard form of the binomial equation $a{{x}^{2}}+bx+c=0$.
Thus,
$\begin{align}
& a=3 \\
& b=-7 \\
& c=1
\end{align}$
Use the quadratic formula, $x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ ,
$\begin{align}
& x=\frac{-\left( -7 \right)\pm \sqrt{{{\left( -7 \right)}^{2}}-4\left( 3 \right)\left( 1 \right)}}{2\left( 3 \right)} \\
& =\frac{7\pm \sqrt{49-12}}{6} \\
& =\frac{7\pm \sqrt{37}}{6}
\end{align}$
Further simplification gives:
$x=\frac{7+\sqrt{37}}{6}$ or $x=\frac{7-\sqrt{37}}{6}$.
Check:
Substitute $x=\frac{7+\sqrt{37}}{6}=2.18$ in the original equation $3{{x}^{2}}-7x+1=0$.
$\begin{align}
3{{\left( 2.18 \right)}^{2}}-7\left( 2.18 \right)+1\overset{?}{\mathop{=}}\,0 & \\
14.26-15.26+1\overset{?}{\mathop{=}}\,0 & \\
14.26-14.26\overset{?}{\mathop{=}}\,0 & \\
0=0 & \\
\end{align}$
Thus, verified.
Substitute $x=\frac{7-\sqrt{37}}{6}=0.152$ in the original equation $3{{x}^{2}}-7x+1=0$.
$\begin{align}
3{{\left( 0.152 \right)}^{2}}-7\left( 0.152 \right)+1\overset{?}{\mathop{=}}\,0 & \\
0.064-1.064+1\overset{?}{\mathop{=}}\,0 & \\
0.064-0.064\overset{?}{\mathop{=}}\,0 & \\
0=0 & \\
\end{align}$
Thus, verified.
Hence, the solution set is $\left\{ \frac{7+\sqrt{37}}{6},\frac{7-\sqrt{37}}{6} \right\}$.
