I am in 6th grade, and I was given this extension for extra practice. The question is:
The ratio of one base to the height to the other base of a trapezoid is $2:3:4$. The area of the trapezoid is 117 square cm. Find the length of the bases and the height.
So far I did $A = \frac{a+b}{2}h \Rightarrow 117=\frac{a+2x}{2}3x$ which gets me to this equation $2x^2+ax-78=0$. Now I am stuck as I don't know how to solve for $a$ or $x$, but can someone nudge me in the right direction?
I edited your question a little, so that it is more readable now.
As you are given the ratios, you should let the common ratio to be the same.
We will denote the length of bases by $b$ and $a$ and the height by $h$. If you let $b = 2x$, then you get $h = 3x$ and $a = 4x$. You have made a slight error of not expressing $a$ in terms of $x$.
Now you should be able to solve for $x$ (which should come out to be $\sqrt{13}$, I believe)