Question 124353
note: I'm going to use x instead of t





{{{y=x^2-12 x+20}}} Start with the given equation



{{{y-20=x^2-12 x}}}  Subtract {{{20}}} from both sides



Take half of the x coefficient {{{-12}}} to get {{{-6}}} (ie {{{(1/2)(-12)=-6}}}).


Now square {{{-6}}} to get {{{36}}} (ie {{{(-6)^2=(-6)(-6)=36}}})





{{{y-20=x^2-12x+36-36}}} Now add and subtract this value inside the parenthesis. Doing both the addition and subtraction of {{{36}}} does not change the equation




{{{y-20=(x-6)^2-36}}} Now factor {{{x^2-12x+36}}} to get {{{(x-6)^2}}}




{{{y=(x-6)^2-36+20}}} Now add {{{20}}} to both sides to isolate y



{{{y=(x-6)^2-16}}} Combine like terms



Now we're done with completing the square



{{{(x-6)^2-16=0}}} Now to solve for x, let {{{y=0}}}



{{{(x-6)^2=+16}}} Add {{{16}}} to both sides




{{{(x-6)^2=16}}} Reduce



*[Tex \LARGE x-6=\pm \sqrt{16}] Take the square root of both sides



*[Tex \LARGE x-6=\pm 4] Simplify the square root



*[Tex \LARGE x=6\pm 4] Add {{{6}}} to both sides



So the solution breaks down to 



{{{x=6+4}}} or {{{x=6-4}}}


Combine like terms


{{{x=10}}} or {{{x=2}}}



So our answers are 



{{{x=10}}} or {{{x=2}}}




Notice if we graph {{{y=x^2-12 x+20}}},  we can see that the roots are {{{x=10}}} and {{{x=2}}}. So this visually verifies our answer.


{{{graph(500,500,-10,12,-10,12,1x^2-12x+20)}}}