Question 589596
solving these equations. 
(3x-7)(x+5) = 0,
Two solutions
x = -5
and
3x = 7
x = 7/3
: 
 w^2 = 16w,
divide both sides by w, results:
w = 16
w = {{{sqrt(16)}}}
w = -4
and
w = 4, this is the only valid solution
: 
 b^2 + 6b - 16 = 0.
Factors to
(b+8)(b-2) = 0; You check this by FOILing these factors
two solutions
b = -8 
and
b = 2