A math problem (algebra) any solution?!1

"Niles and Frasier are brothers who have competed with each other all their lives, and continue to do so as adults. When Niles decided to put a rectangular pool in his backyard this summer, Frasier learned of his plans and decided he had to put a pool in his yard, too. To annoy Niles as much as possible, Frasier plans to build the same shape rectangle as his brother (the ratio of length to width is identical), but of course he will make his larger.

When Niles proudly announced his plans to Frasier, he said, "My new pool will be 4 meters wide."

"That's nothing!" exclaimed Frasier. "I'm building one, too, and my new pool will be 6 meters longer than yours. And the width of my pool will be the same as the length of yours!"

Find the area of Frasier's new pool. Express your answer in simplified radical form - don't use any rounded decimals.

Note: One easy way to write radicals on a keyboard is to use SQRT( ), so SQRT(25) = 5.

Extra: The average depth of Frasier's new pool will be 2 meters. Find the average depth Niles would need in his pool for the two pools to hold the same amount of water. Again, give your answer in radical form"


(from problem of the day)

LEGEND
W= WIDTH
L= LENGTH
D= DEPTH
A= AREA

NILE'S POOL DIMENSIONS
WN= 4
LN= L
DN= D
FRASIER'S POOL DIMENSIONS
WF= L
LF= L+6
DF= 2

THE RATIO LENGTH TO WIDTH IS IDENTICAL
L/(L+6) = 4/L
L2-4L-24=0
L = (4+SQRT(4^2-4*1*(-24)))/2 = 7.29 m
OR
L = (4-SQRT(4^2-4*1*(-24)))/2 = -3.291502622m NOT ACCEPTABLE
L= 7.29 m
IN A RADICAL FORM
L= 2+2*SQRT 7

AREA OF FRASIER'S NEW POOL= L^2+6*L
AF= 96.92 m^2

THE TWO POOLS HOLD THE SAME AMOUNT OF WATER
WN*LN*DN = WF*LF*DF
DN= WF*LF*DF/WN/LN
DN= 6.65 m

AMOUNT OF WATER IN EACH POOL= 193.83 m^3
RATIO LENGTH TO WIDTH FOR EACH POOL = 1.82

SORRY, ONLY FIRST ANSWER IS IN A RADICAL FORM