<s>[INST] Solve for $+r$ ; $A=2\pi r^2+2\pi rh$ 
Solve for $+r$

$$A=2\pi r^2+2\pi rh$$ Since $2\pi$ is common on both sides of the $+$ so I will take it out
$$A=2\pi (r^2+rh)$$ 
Now, divide both sides by $2\pi$
$$\dfrac{A}{2\pi}=r^2+rh$$
Then, we can divide by the $h$
$$\dfrac{Ah}{2\pi}=r^2+r$$
Then; 
$$0=r^2+r-\dfrac{Ah}{2\pi}$$ 
Quadratic formula $=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ $a=1,b=1,c=?$ What are the values of $a,b,c$
 [/INST] No! $\sqrt{r^2 + r} \neq 2r.$
What you have is a polynomial equation $r^2 + r - \dfrac{Ah}{\pi} = 0.$
Since this is a homework, I will only give a hint: the two roots $r_1, r_2$ of quadratic $ax^2 + bx + c = 0$ are $$ r_{1,2} = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}.$$
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