3 yards greater than the width = w+3
where
width = w and length = L
However, since w+3 = L, we substitute w+3 for L.
Area of a rectangle: wL, or width x length.
Substitute: w x (w+3) = 270
Now, just solve.
We start by distributing the left side of the equation.
w(w+3) = w^2 + 3w = 270
Subtract 270 from both sides.
w^2 +3w - 270 = 0
We can't FOIL here, so back to good ol' quadratic formula.
Formula: (-b ± √(b²-4ac))/2a, where:
aw^2 + bw - c = 0
Thus:
a=1, b=3, c=-270.
Plug these values in for the variables in the equation, and you get two different values (depending on the plus/minus)
15 and -18.
However, distance(which is essentially the length of a certain line or how long something is) cannot be negative, so the only answer is 15.
Width is 15.
Plug into our original equation, 15 for w.
15L=270
Divide by 15 on both sides.
L = 18
Width: 15
Length: 18
