legnth is 3 more than twice its width l = 3 + 2 times w l=3+2w a=65 a=l times w subsitute 65=lw l=3+2w subsitute 65=(3+2w)(w) 65=3w+2w^2 subtract 65 from both sides 0=2w^2+3w-65
if we can factor then we can do if zy=0 then assume z and/or y=0 so to factor in ax^2+bx+c form when a is greater than 1,
aw^2+bw+c 2=a 3=b -65=c
to factor first find a times c=z b=t+y t times y=z so a times c=2 times -65=-130 b=3
factor -130 -130= -1,130 -2,65 -5,26 -10,13
now add them together and see which ones make 3 or -3 (b)
-1+130=129 -2+65=63 -5+26=21 -10+13=3 match
seperate the middle number like that 3w=-10w+13w
so 2w^2-10w+13w-65=0 group (2w^2-10w)+(13w-65)=0 factor (2w)(w-5)+(13)(w-5)=0 reverse distribute ab+ac=a(b+c) (2w+13)(w-5)=0 set each to zero 2w+13=0 w-5=0
solve 2w+13=0 subtract 13 2w=-13 divide 2 w=-6 and 1/2 impossible since width cannot be negative discard