# Trouble factoring polynomial

So I've gotten pretty far on this problem but I'm unable to complete it. I don't want to resort to the quadratic formula because this section of my review sheet specifically tells me not to use it. The problem is as follows:

$$3e^{2t} – e^t = 70$$

I substitute $y$ for $e^t$, and divide by 3, rewriting the equation is:

$$y^2 - \frac{y}{3} = \frac{70}{3}$$

But now I'm stuck. Factoring by grouping doesn't work and it doesn't seem like you can complete the square, could somebody help?

• Why can't you complete the square? – Eric Towers Jun 4 '14 at 21:32
• Completing the square actually works fine. You get a nice square on the right. – Daniel Fischer Jun 4 '14 at 21:33
• You can always complete the square! :) – Just_a_fool Jun 4 '14 at 21:35

$$3y^2-y-70=0$$
Now look for two numbers whose product is $(3)(-70) = -210$ and whose sum is $-1$ (the latter from the coefficient of the $y$ term).
Those numbers must be nearly equal if their sum is $-1$, so they must be close to $\sqrt{210}$ in absolute value, which is around $14$ or $15$. And lo! the numbers $14$ and $-15$ fit the bill. Next:
$$3y^2 + 14y -15y -70 = 0$$ $$y(3y+14) - 5(3y+14) = 0$$ $$(y-5)(3y+14)=0$$
$$y^2 - \frac{y}{3} + \left( \frac{1}{2 \cdot 3} \right)^2 = \left(y - \frac{1}{6} \right)^2$$ $$\frac{70}{3} + \left( \frac{1}{2 \cdot 3} \right)^2 = \left( \frac{29}{6} \right)^2$$