3x+x^2=160 find x rearrangimg by subtracting 160 from both sides, we have
x^2 + 3x - 160 = 0 this can't be factored....so using the on-site solver, we have
$${{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{160}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{649}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}\right)}{{\mathtt{2}}}}\\
{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{649}}}}{\mathtt{\,-\,}}{\mathtt{3}}\right)}{{\mathtt{2}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{14.237\: \!739\: \!202\: \!856\: \!996\: \!8}}\\
{\mathtt{x}} = {\mathtt{11.237\: \!739\: \!202\: \!856\: \!996\: \!8}}\\
\end{array} \right\}$$
Rewrite 3x + x² = 160 as x² + 3x - 160 = 0
Use the quadratic formula: x = ( -b ± √( b² - 4·a·c ) / ( 2·a )
with a = 1 b = 3 and c = -160.
3x+x^2=160 find x rearrangimg by subtracting 160 from both sides, we have
x^2 + 3x - 160 = 0 this can't be factored....so using the on-site solver, we have
$${{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{160}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{649}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}\right)}{{\mathtt{2}}}}\\
{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{649}}}}{\mathtt{\,-\,}}{\mathtt{3}}\right)}{{\mathtt{2}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{14.237\: \!739\: \!202\: \!856\: \!996\: \!8}}\\
{\mathtt{x}} = {\mathtt{11.237\: \!739\: \!202\: \!856\: \!996\: \!8}}\\
\end{array} \right\}$$