Compute the perimeter of a right-angled triangle with area 7 and hypotenuse of length 6.?

Let a and b be the two sides of the triangle.

Area of the triangle :
(1/2)ab = 7
2ab = 28 ...... [1]

By Pythagorean theorem :
√(a² + b²) = 6
a² + b² = 36 ...... [2]

[1] + [2] :
a² + 2ab + b² = 64
(a + b)² = 64
a + b = 8

Perimeter
= a + b + 6
= 8 + 6
= 14

ab/2 = 7, so ab = 2*7 = 14, so b = 14/a
a^2 + b^2 = 6^2
a^2 + (14/a)^2 = 36
a^2 + (196/(a^2)) - 36 = 0
a^4 - 36a^2 + 196 = 0
Let x = a^2, so we have:
x^2 - 36x + 196 = 0
x = (-(-36) +/- sqrt((-36)^2 - 4*1*196)) / (2*1)
x = (36 +/- sqrt(1,296 - 784)) / 2
x = (36 +/- sqrt(512)) / 2
x = 18 +/- sqrt(128)
a = sqrt(18 + sqrt(128)) and b = sqrt(18 - sqrt(128)), or vice versa.
Either way the perimeter is:
sqrt(18 + sqrt(128)) + sqrt(18 - sqrt(128)) + 6 = 14

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