Question 353843
Find the area of a right triangle whose legs have lengths which differ by 7 meters and whose hypotenuse is 17 meters long. (Show your solution using Quadratic Equation)
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Once you see "right triangle" you can apply the Pythagorean theorem.
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Let x = shortest leg
then
x+7 = other leg
17 = hypotenuse
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x^2 + (x+7)^2 = 17^2

x^2 + (x+7)(x+7) = 17^2

x^2 + x^2+14x+49 = 289

2x^2+14x+49 = 289

2x^2+14x-240 = 0

x^2+7x-120 = 0 (your quadratic)

(x+15)(x-8) = 0

x = {-15, 8}

Throw out the negative solution leaving
x = 8 meters (shortest leg)
x+7 = 8+7= 15 meters (other leg)


Area is 
(1/2)bh
(1/2)(8)(15)
(4)(15)
60 square meters