## Answer :

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.

Length= 2w+3

(Length)(Width)= Area

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.