Factoring an unusual polynomial

Question

Hello I was presented today with a logarithmic problem. After computing the calculations in order to solve for $x$ we had to factor the following equation (no calculator allowed). I am a somewhat stumped due to the numbers. Is there an ingenious way of solving it?

$35^3$ = $x^2 + 70x$

Answer 1

You have

$$x^2 + 70x - 35^3 = 0 \label{eq1A}$$

Using the quadratic formula, you can get

$$\begin{equation}\begin{aligned} x & = \frac{-70 \pm \sqrt{70^2 + 4(35^3)}}{2} \\ & = \frac{-70 \pm \sqrt{70^2 + (2)(35)(2)(35)(35)}}{2} \\ & = \frac{-70 \pm \sqrt{70^2(1 + 35)}}{2} \\ & = \frac{-70 \pm 70(6)}{2} \\ & = 35(-1 \pm 6) \\ & = -245 \; \text{ or } \; 175 \end{aligned}\end{equation}\tag{2}\label{eq2A}$$

Answer 2

After rearranging, you get a quadratic. Note that the discriminant of this quadratic is $ 70^2 + 4(1)(35^3) = 4(35)^2(1+35)= (2^2)(35^2)(6^2)$, a perfect square.