+118651 3x^2+2x+1=0
lets look at the discriminate in the quadratic formula
$$\triangle =b^2-4ac\\\\
\triangle =2^2-4*3*1\\\\
\triangle =4-12\\\\
\triangle =-8\\\\$$
In the full formula you have to take the sqrt of this and you cannot take the sqrt of a negative number.
Therefore this has no solutions (it has imaginary ones but you probably don't need to worry about those. )
+33653 There are two complex solutions. These can be found by using the quadratic formula or just enter them in the calculator on the home page here.
$${\mathtt{3}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{2}}}}{\mathtt{\,\times\,}}{i}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{{\mathtt{3}}}}\\
{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{2}}}}{\mathtt{\,\times\,}}{i}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{{\mathtt{3}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.471\: \!404\: \!520\: \!791\: \!189\: \!8}}{i}\right)\\
{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{3}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.471\: \!404\: \!520\: \!791\: \!189\: \!8}}{i}\\
\end{array} \right\}$$
+118651 3x^2+2x+1=0
lets look at the discriminate in the quadratic formula
$$\triangle =b^2-4ac\\\\
\triangle =2^2-4*3*1\\\\
\triangle =4-12\\\\
\triangle =-8\\\\$$
In the full formula you have to take the sqrt of this and you cannot take the sqrt of a negative number.
Therefore this has no solutions (it has imaginary ones but you probably don't need to worry about those. )
