Question 251684
{{{(3w-2)(w+5)=90}}}
Then I foiled and got:
{{{3w^2+13w-100=0}}}
And I get stuck there :/ I don't think it factors and I'm not sure if I foiled wrong or set it up wrong. :/ Help me, please? Thanks :)
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This is all good. If you think this trinomial does not factor, then use the quadratic formula on it. (This trinomial actually does factor but the quadratic formula still works.)
{{{w = (-b +- sqrt(b^2 - 4ac))/2a}}}
Your a = 3, b = 13 and c = -100:
{{{w = (-(13) +- sqrt((13)^2 - 4(3)(-100)))/2(3)}}}
{{{w = (-13 +- sqrt(169 - 4(3)(-100)))/6}}}
{{{w = (-13 +- sqrt(169 + 1200))/6}}}
{{{w = (-13 +- sqrt(1369))/6}}}
{{{w = (-13 +- 37)/6}}}
{{{w = (-13 + 37)/6}}} or {{{w = (-13 - 37)/6}}}
{{{w = 24/6}}} or {{{w = (-50)/6}}}
{{{w = 4}}} or {{{w = (-25)/3}}}
Since w represents the side of a rectangle we will reject the negative solution. So the only possible value for w that works is w = 4. And that makes the original length, which is 3w, 12.<br>
BTW: {{{3w^2+13w-100=0}}} factors into:
{{{(3w + 25)(w - 4) = 0}}}
And this gives us the same answers as the quadratic formula did.