Question 174882
{{{t = 200 - 5x - x^2}}} Start with the given equation.



{{{t =  - x^2- 5x+200 }}} Rearrange the terms.



{{{50 =  - x^2- 5x+200 }}} Plug in {{{t=50}}}



{{{0 =  - x^2- 5x+200-50 }}} Subtract 50 from both sides.



{{{0 =  - x^2- 5x+150 }}} Combine like terms.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=-1}}}, {{{b=-5}}}, and {{{c=150}}}



Let's use the quadratic formula to solve for x



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



{{{x = (-(-5) +- sqrt( (-5)^2-4(-1)(150) ))/(2(-1))}}} Plug in  {{{a=-1}}}, {{{b=-5}}}, and {{{c=150}}}



{{{x = (5 +- sqrt( (-5)^2-4(-1)(150) ))/(2(-1))}}} Negate {{{-5}}} to get {{{5}}}. 



{{{x = (5 +- sqrt( 25-4(-1)(150) ))/(2(-1))}}} Square {{{-5}}} to get {{{25}}}. 



{{{x = (5 +- sqrt( 25--600 ))/(2(-1))}}} Multiply {{{4(-1)(150)}}} to get {{{-600}}}



{{{x = (5 +- sqrt( 25+600 ))/(2(-1))}}} Rewrite {{{sqrt(25--600)}}} as {{{sqrt(25+600)}}}



{{{x = (5 +- sqrt( 625 ))/(2(-1))}}} Add {{{25}}} to {{{600}}} to get {{{625}}}



{{{x = (5 +- sqrt( 625 ))/(-2)}}} Multiply {{{2}}} and {{{-1}}} to get {{{-2}}}. 



{{{x = (5 +- 25)/(-2)}}} Take the square root of {{{625}}} to get {{{25}}}. 



{{{x = (5 + 25)/(-2)}}} or {{{x = (5 - 25)/(-2)}}} Break up the expression. 



{{{x = (30)/(-2)}}} or {{{x =  (-20)/(-2)}}} Combine like terms. 



{{{x = -15}}} or {{{x = 10}}} Simplify. 



So the possible answers are {{{x = -15}}} or {{{x = 10}}} 

  

Since "x" is the position, this means that a negative position doesn't make sense. So the value {{{x = -15}}} can be ignored.



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Answer:


So the solution is {{{x = 10}}} which means that at 10 units (or whatever units you're dealing with), the temperature will be 50 degrees.