algebra/geometry

$$x(x+29)=1750$$

$$x^2+29x-1750=0$$

now use the quadratic formula to find the values of x that satisfy this

a=1

b=29

c=-1750

$$r1,r2=\dfrac{-b\pm \sqrt{b^2-4 a c}}{2 a}$$

$${\frac{\left({\mathtt{\,-\,}}{\mathtt{29}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{{\mathtt{29}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{1}}{\mathtt{\,\times\,}}\left(-{\mathtt{1\,750}}\right)}}\right)}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{1}}\right)}} = {\mathtt{29.774\: \!710\: \!614\: \!525\: \!76}}$$

$${\frac{\left({\mathtt{\,-\,}}{\mathtt{29}}{\mathtt{\,-\,}}{\sqrt{{{\mathtt{29}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{1}}{\mathtt{\,\times\,}}\left(-{\mathtt{1\,750}}\right)}}\right)}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{1}}\right)}} = -{\mathtt{58.774\: \!710\: \!614\: \!525\: \!76}}$$

and these are the two values of x that satisfy the original equation

$${\mathtt{29.774\: \!7}}{\mathtt{\,\times\,}}\left({\mathtt{29.774\: \!7}}{\mathtt{\,\small\textbf+\,}}{\mathtt{29}}\right) = {\mathtt{1\,749.999\: \!060\: \!09}}$$

$$\left(-{\mathtt{58.774\: \!7}}{\mathtt{\,\times\,}}\left({\mathtt{\,-\,}}{\mathtt{58.774\: \!7}}{\mathtt{\,\small\textbf+\,}}{\mathtt{29}}\right)\right) = {\mathtt{1\,749.999\: \!060\: \!09}}$$

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