The formula of an area of a triangle with base b and height h:
[tex]A_{\Delta}=\dfrac{bh}{2}[/tex]
We have:
[tex]A_{\Delta}=672\ in^2\\b=5x-4\\h=2x[/tex]
[tex]Domain:\\5x-4 > 0\to x > 0.8\\x > 0\\therefore\ D:x > 0.8[/tex]
Substitute:
[tex]\dfrac{(5x-4)\cdot2x}{2}=672\\\\x(5x-4)=672\ \ \ |-672\\\\5x^2-4x-672=0\\\\5x^2+56x-60x-673=0\\\\x(5x+56)-12(5x+56)=0\\\\(5x+56)(x-12)=0\iff5x+56=0\ \vee\ x-12=0\\\\5x+56=0\ \ \ |-56\\5x=-56\ \ \ |:5\\x=-11.2\notin D\\\\x-12=0\ \ \ |+12\\x=12\in D[/tex]
Answer: x = 12 in.
