SOLUTION: Find the dimensions of a right triangle if the area is 48 and one leg is 4 less than twice the other leg. my attempt goes this way: A=1/2 bh 48=1/2 (2x-4) (x) 48=1/2 (2x^2

Question 986511: Find the dimensions of a right triangle if the area is 48 and one leg is 4 less than twice the other leg.
my attempt goes this way:
A=1/2 bh
48=1/2 (2x-4) (x)
48=1/2 (2x^2-4x)
48=x^2-2x
0=x^2-2x-48
then i used quadratic formula but i'll only use plus sign here because i don't know how to type in the plus and minus sign.
----it seems i can't type the equation properly here--but i only substituted the formula from the equation i got in the area part.

here is my theory or the answer i got.
x=8
2x-4=12
i can't seem to get the hypotenuse using these two answers because i get a decimal answer.
c^2=a^2+b^2 , using the Pythagorean theorem, i got an answer with a decimal, which in my opinion is wrong.
Answer by MathLover1(20849) (Show Source): You can put this solution on YOUR website!
........correct
........correct
........correct
........correct
........correct

positive solution:

=> this is a height
now find the length of the base

=>the base
check:

the hypotenuse:

........must be decimal

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