How do you find the radius of a rod whose surface area is 70pi in^2 and height of 19 in? r=sqr(A/2pi-rh) but what is the answer

The surface area is given by

SA= 2*pi*r*h + 2*pi*r^2      so we have

70 = 2*pi*r*h + 2*pi*r^2     divide by 2pi   and rearrange

r^2 + hr - 35/pi = 0              add 35/pi to both sides

r^2 + hr  =  35/pi                complete the square

r^2 + hr + (h^2/4) = 35/pi + (h^2/4)  factor

(r + h/2 )^2 = 35/pi + (h/2)^2   take the positive root of both sides

r + h/2 = √(35/pi + (h/2)^2)   subtract h/2 from both sides

r = √(35/pi + (h/2)^2) - h/2

You have written your equation in implicit form. i.e. "r" is on both sides of the equation.  Better to write it in explicit form so that you can calculate "r" directly:

$$r=-\frac{h}{2}+\sqrt{(\frac{h}{2})^2+\frac{A}{2\pi}}$$

.

$$\\r=sqr(A/(2\pi)-rh)\\\\
r=\sqrt{\frac{70}{2\pi}-19r}\\\\
r^2=\frac{70}{2\pi}-19r\\\\
2\pi r^2=70-19*2\pi r\\\\
\pi r^2+19\pi r-35=0\\\\$$

$${\mathtt{\pi}}{\mathtt{\,\times\,}}{{\mathtt{r}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{19}}{\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\times\,}}{\mathtt{r}}{\mathtt{\,-\,}}{\mathtt{35}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{r}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{361}}{\mathtt{\,\times\,}}{{\mathtt{\pi}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{140}}{\mathtt{\,\times\,}}{\mathtt{\pi}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{19}}{\mathtt{\,\times\,}}{\mathtt{\pi}}\right)}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{\pi}}\right)}}\\
{\mathtt{r}} = {\frac{\left({\sqrt{{\mathtt{361}}{\mathtt{\,\times\,}}{{\mathtt{\pi}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{140}}{\mathtt{\,\times\,}}{\mathtt{\pi}}}}{\mathtt{\,-\,}}{\mathtt{19}}{\mathtt{\,\times\,}}{\mathtt{\pi}}\right)}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{\pi}}\right)}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{r}} = -{\mathtt{19.569\: \!302\: \!161\: \!343\: \!291}}\\
{\mathtt{r}} = {\mathtt{0.569\: \!302\: \!161\: \!343\: \!291}}\\
\end{array} \right\}$$

SO

$$r\approx 0.57\; inches$$