# The length of a rectangle is 5 yd more than twice its width, and the area of the rectangle is 42yd^2 . How do I find the dimensions of the rectangle?

Sep 27, 2016

Let the length be $2 x + 5$ and the width be $x$.

$x \left(2 x + 5\right) = 42$

$2 {x}^{2} + 5 x = 42$

$2 {x}^{2} + 5 x - 42 = 0$

$2 {x}^{2} + 12 x - 7 x - 42 = 0$

$2 x \left(x + 6\right) - 7 \left(x + 6\right) = 0$

$\left(2 x - 7\right) \left(x + 6\right) = 0$

$x = \frac{7}{2} \mathmr{and} - 6$

Hence, the dimensions are $\frac{7}{2}$ by $12$ yards.

Hopefully this helps!