Twice the width of a rectangle is 3 m more than the length. If the area of the rectangle is209 m2, find the dimensions of the rectangle.

Sure, let's solve this step by step.

Step 1: Let's denote the width of the rectangle as W and the length as L.

Step 2: According to the problem, twice the width of the rectangle is 3 m more than the length. We can write this as an equation: 2W = L + 3.

Step 3: We also know that the area of the rectangle is 209 m^2. The area of a rectangle is calculated by multiplying its length by its width, so we can write this as another equation: L * W = 209.

Step 4: Now we have a system of two equations, and we can solve it. Let's substitute the first equation (2W = L + 3) into the second one (L * W = 209). This gives us: W * (2W - 3) = 209.

Step 5: This is a quadratic equation in terms of W. We can rewrite it as: 2W^2 - 3W - 209 = 0.

Step 6: Now we can solve this quadratic equation to find the value of W. The solutions of the quadratic equation ax^2 + bx + c = 0 are given by the formula: x = [-b ± sqrt(b^2 - 4ac)] / (2a). In our case, a = 2, b = -3, and c = -209.

Step 7: Substituting these values into the formula gives us: W = [3 ± sqrt((-3)^2 - 42(-209))] / (2*2) = [3 ± sqrt(9 + 1672)] / 4 = [3 ± sqrt(1681)] / 4 = [3 ± 41] / 4.

Step 8: This gives us two possible solutions: W = 44/4 = 11 m or W = -38/4 = -9.5 m. Since the width cannot be negative, we discard the second solution.

Step 9: Substituting W = 11 m into the first equation (2W = L + 3) gives us: L = 2W - 3 = 2*11 - 3 = 19 m.

So, the dimensions of the rectangle are 11 m (width) and 19 m (length).