Question 167792
Note: I'm replacing "l" with "x" (since "l" is often confused with 1)



{{{(x + 2) (x-4) = 20}}} Start with the given equation



{{{x^2-2x-8 = 20}}} FOIL



{{{x^2-2x-8 - 20=0}}} Subtract 20 from both sides.



{{{x^2-2x-28=0}}} Combine like terms.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=1}}}, {{{b=-2}}}, and {{{c=-28}}}



Let's use the quadratic formula to solve for x



{{{x = (-b +- sqrt( b^2-4ac ))/(2a)}}} Start with the quadratic formula



{{{x = (-(-2) +- sqrt( (-2)^2-4(1)(-28) ))/(2(1))}}} Plug in  {{{a=1}}}, {{{b=-2}}}, and {{{c=-28}}}



{{{x = (2 +- sqrt( (-2)^2-4(1)(-28) ))/(2(1))}}} Negate {{{-2}}} to get {{{2}}}. 



{{{x = (2 +- sqrt( 4-4(1)(-28) ))/(2(1))}}} Square {{{-2}}} to get {{{4}}}. 



{{{x = (2 +- sqrt( 4--112 ))/(2(1))}}} Multiply {{{4(1)(-28)}}} to get {{{-112}}}



{{{x = (2 +- sqrt( 4+112 ))/(2(1))}}} Rewrite {{{sqrt(4--112)}}} as {{{sqrt(4+112)}}}



{{{x = (2 +- sqrt( 116 ))/(2(1))}}} Add {{{4}}} to {{{112}}} to get {{{116}}}



{{{x = (2 +- sqrt( 116 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (2 +- 2*sqrt(29))/(2)}}} Simplify the square root  (note: If you need help with simplifying square roots, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)  



{{{x = (2)/(2) +- (2*sqrt(29))/(2)}}} Break up the fraction.  



{{{x = 1 +- 1*sqrt(29)}}} Reduce.  



{{{x = 1+sqrt(29)}}} or {{{x = 1-sqrt(29)}}} Break up the expression.  



So the possible answers are {{{x = 1+sqrt(29)}}} or {{{x = 1-sqrt(29)}}} 



which approximate to {{{x=6.385}}} or {{{x=-4.385}}} 



Since we're looking for a measurement, and a negative measurement doesn't make much sense, this means that the only answer is {{{x = 1+sqrt(29)}}} which approximates to {{{x=6.385}}}