Question 624615
Let the length of the triangle be x.

width= x + 8

area= x * (x+8)
    =x(x+8)
    =x^2 + 8x

280=x^2 + 8x
x^2 + 8x - 280= 0 <====quadratic equation

{{x = (-8 +- sqrt( b^2-4*1*-280 ))/(2*1) }}

{{{x = (-8 +- 4sqrt( 74 ))/(2) }}} 

x=-4(+/-)2sqrt(74)

x=13.2 (nearest decimal place)
or
x=-21.2

The width can't be a negative because you can't have a negative width.
Therefore, the width is 13.2 to the nearest decimal place.