Answer:
![Perimeter=54ft^2](https://img.qammunity.org/2020/formulas/mathematics/middle-school/th7iduvhep63k01gennpdnuw1p37mukoe2.png)
Explanation:
Let's start by writing what we know in mathematical terms.
A rectangle has a length that is 3 more than twice its width, means:
![L-3 = 2W\\L*W = 152](https://img.qammunity.org/2020/formulas/mathematics/middle-school/b5mvqgezxt4h8v6zj7qif6b8llwe9cdef3.png)
Now all we have to do is solve for one of these values. I'll choose W.
Rewrite the equation:
![L = 2W + 3](https://img.qammunity.org/2020/formulas/mathematics/middle-school/svp7fok3nlgc7lh2nr8ee2j5nypdfccyv4.png)
Input that information into our second equation and solve:
![W*(2W + 3) = 152\\2W^2+3W=152\\2W^2+3W-152=0](https://img.qammunity.org/2020/formulas/mathematics/middle-school/wnaf0g65jruugo76wck93xjez38irjhb2r.png)
Find the two W values by factoring the equation (note that the Width can't be negative because you can't have a negative rectangle):
![2W^2+3W-152=0\\(2W+19)(W-8)=0\\\\W=-(19)/(2)\\W=8](https://img.qammunity.org/2020/formulas/mathematics/middle-school/c7m6kp7fkzcp51okl6nsgj57b58p2rc0c6.png)
We pick the positive value 8 and plug it into our original equation to find Length:
![L = 2W + 3\\L=2(8)+3\\L=19](https://img.qammunity.org/2020/formulas/mathematics/middle-school/j37va3lxtuy3ekrc57j8bvadyjlbqvht0p.png)
Finally, after all that, we can use the formula of a perimeter using our newly found Length and Width values (remember not to forget "ft^2" when we get our answer):
![2L+2W\\2(19)+2(8)\\Perimeter=54ft^2](https://img.qammunity.org/2020/formulas/mathematics/middle-school/farpafbyvogoqom62armm298z8m4hwb9zk.png)