A rectangle has a length that is 3 feet more than the width. If the area of the rectangle is 52 square feet more than the area of a square with a side of 6, what are the dimensions of the rectangle?

Step-by-step explanation:

given:

length of rectangle=3+w

area of rectangle=58+area of square

side of square=6

SOLUTION

area of square=L^2

area of square=6^2

area of square = 36

area of rectangle=58+area of square

area of rectangle=58+36

=94

area of rectangle = L*w

94 = (3+w) * w

94=3w+w^2

w^2+3w=94

w^2+3w + (3/2) ^2=94 + (3/2) ^2

(w+3/2) ^2=94+9/4

(w+3/2) ^2=385/4

√ (w+3/2) ^2=√385/√4

(w+3/2) = ±√385/2

hope it gives you a hint