Question 92081
I'll pick up where you left off (your steps are correct)


{{{2x^2 -12x + 20 = 100 }}}


{{{2x^2 -12x + 20 - 100=0 }}} Subtract 100 from both sides


{{{2x^2 -12x -80=0 }}} Combine like terms



Now let's use the quadratic formula to solve for x:



Starting with the general quadratic


{{{ax^2+bx+c=0}}}


the general solution using the quadratic equation is:


{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a)}}}


So lets solve {{{2*x^2-12*x-80=0}}} ( notice {{{a=2}}}, {{{b=-12}}}, and {{{c=-80}}})


{{{x = (--12 +- sqrt( (-12)^2-4*2*-80 ))/(2*2)}}} Plug in a=2, b=-12, and c=-80




{{{x = (12 +- sqrt( (-12)^2-4*2*-80 ))/(2*2)}}} Negate -12 to get 12




{{{x = (12 +- sqrt( 144-4*2*-80 ))/(2*2)}}} Square -12 to get 144  (note: remember when you square -12, you must square the negative as well. This is because {{{(-12)^2=-12*-12=144}}}.)




{{{x = (12 +- sqrt( 144+640 ))/(2*2)}}} Multiply {{{-4*-80*2}}} to get {{{640}}}




{{{x = (12 +- sqrt( 784 ))/(2*2)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (12 +- 28)/(2*2)}}} Simplify the square root (note: If you need help with simplifying the square root, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)




{{{x = (12 +- 28)/4}}} Multiply 2 and 2 to get 4


So now the expression breaks down into two parts


{{{x = (12 + 28)/4}}} or {{{x = (12 - 28)/4}}}


Lets look at the first part:


{{{x=(12 + 28)/4}}}


{{{x=40/4}}} Add the terms in the numerator

{{{x=10}}} Divide


So one answer is

{{{x=10}}}




Now lets look at the second part:


{{{x=(12 - 28)/4}}}


{{{x=-16/4}}} Subtract the terms in the numerator

{{{x=-4}}} Divide


So another answer is

{{{x=-4}}}


So our possible solutions are:

{{{x=10}}} or {{{x=-4}}}






Now lets find each of the leg's lengths:

Leg A: 
{{{10-2=8}}} Plug in x=10

{{{-4-2=-6}}} Plug in x=-4

Since a negative length doesn't make sense, the solution x=-4 must be discarded



Leg B: 
{{{10-4=6}}} Plug in x=10

{{{-4-4=-8}}} Plug in x=-4

Since a negative length doesn't make sense, the solution x=-4 must be discarded



So the only solution is 


{{{x=10}}}  where the lengths of the legs are 8 and 6 (or 6 and 8)