Question 571712
So as not to repeat the whole words "length" and "width" too many times, we abbreviate by saying
Let L be the length, and let W be the width.
{{{L=W+30}}}
{{{Area=L*W}}} --> {{{Area=(W+30)W}}}
We get the equation
{{{(W+30)W=5000}}} --> {{{W^2+30W=5000}}} --> {{{W^2+30W-5000=0}}}
We can find the solutions to that equation, applying the quadratic formula, as
{{{W = (-30 +- sqrt( 30^2-4*1*(-5000)))/(2*1)=(-30 +- sqrt(900+20000))/2=(-30 +- sqrt(20900))/2=(-30 +- sqrt(100*209))/2=(-30 +- sqrt(100)*sqrt(209))/2=(-30 +- 10sqrt(209))/2 =2(-15 +- 5sqrt(209))/2=-15 +- 5sqrt(209)}}}
We need to discard the negative solution, because the width is a positive number, and we are left with {{{highlight(W=-15 + 5sqrt(209))}}}
{{{L=-15 + 5sqrt(209)+30}}} --> {{{highlight(L=15 + 5sqrt(209))}}}
Those values are not pretty, but are exact solutions. They cannot be written as exact decimals, but W = approximately 57.284 and L = approximately 87.284