3x+x^2=160 find x

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.