A=l*w if the length is 3cm more than the width let the length be x+3 and let the width be x since the area=90cm^2 90=(x+3) *x 90=x(x+3) 90=x²+3x x²+3x-90=0 x=(-3+ or - √369) /2 x= 8.105 and -11.105 but the dimensions must be positive so they are 8.105 and 11.105 (they are rounded to 3 decimal places, if you don't want them rounded look at them in the calculator.
w(2w+3) 2w^2+3w= 90 2w^2+3w-90= 0 (2w+15)(w-6)=0 This can be broken down into two different equations: 2w+15=0 ---------------> w= -15/2 w-6=0 -------------------> w= 6
Since w cannot be negative, the width is 6 cm, and the length is 15 cm.
This can also be solved using the quadratic formula: -b±√[b^2-4(a)(c)] 2a
Start with 2w^2+3w-90= 0 a= 2 b= 3 c= -90 -3±√[9-4(2)(-90)] 4 -3±√(729) 4 -3±27 4 Therefore, the two answers are 6, and -30/4, AKA -15/2. Once again, 6 is the only one that works because it is positive, so when plugged in, the length is 15, and the width is 6.
