Question 112305
{{{y^2-12y+36=49}}} Start with the given equation



{{{y^2-12y-13=0}}} Subtract 49 from both sides


{{{(y-13)(y+1)=0}}} Factor the left side (note: if you need help with factoring, check out this <a href=http://www.algebra.com/algebra/homework/playground/change-this-name4450.solver>solver</a>)




Now set each factor equal to zero:


{{{y-13=0}}} or {{{y+1=0}}}


{{{y=13}}} or {{{y=-1}}}  Now solve for y in each case



So our solutions are {{{y=13}}} or {{{y=-1}}}



Notice if we graph {{{y=x^2-12x-13}}} (just replace y with x) we get


{{{ graph(500,500,-10,15,-10,10, x^2-12x-13) }}}


and we can see that the graph has roots at {{{y=13}}} and {{{y=-1}}}, so this verifies our answer.




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#2



{{{c^2+16c+64=15}}} Start with the given equation



{{{c^2+16c+49=0}}} Subtract 15 from both sides



Let's use the quadratic formula to solve for c:



Starting with the general quadratic


{{{ac^2+bc+c=0}}}


the general solution using the quadratic equation is:


{{{c = (-b +- sqrt( b^2-4*a*c ))/(2*a)}}}




So lets solve {{{c^2+16*c+49=0}}} ( notice {{{a=1}}}, {{{b=16}}}, and {{{c=49}}})





{{{c = (-16 +- sqrt( (16)^2-4*1*49 ))/(2*1)}}} Plug in a=1, b=16, and c=49




{{{c = (-16 +- sqrt( 256-4*1*49 ))/(2*1)}}} Square 16 to get 256  




{{{c = (-16 +- sqrt( 256+-196 ))/(2*1)}}} Multiply {{{-4*49*1}}} to get {{{-196}}}




{{{c = (-16 +- sqrt( 60 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{c = (-16 +- 2*sqrt(15))/(2*1)}}} Simplify the square root (note: If you need help with simplifying the square root, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)




{{{c = (-16 +- 2*sqrt(15))/2}}} Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


{{{c = (-16 + 2*sqrt(15))/2}}} or {{{c = (-16 - 2*sqrt(15))/2}}}



Now break up the fraction



{{{c=-16/2+2*sqrt(15)/2}}} or {{{c=-16/2-2*sqrt(15)/2}}}



Simplify



{{{c=-8+sqrt(15)}}} or {{{c=-8-sqrt(15)}}}



So these expressions approximate to


{{{c=-4.12701665379258}}} or {{{c=-11.8729833462074}}}



So our solutions are:

{{{c=-4.12701665379258}}} or {{{c=-11.8729833462074}}}


Notice when we graph {{{x^2+16*x+49}}} (just replace c with x), we get:


{{{ graph( 500, 500, -21.8729833462074, 5.87298334620742, -21.8729833462074, 5.87298334620742,1*x^2+16*x+49) }}}


when we use the root finder feature on a calculator, we find that {{{x=-4.12701665379258}}} and {{{x=-11.8729833462074}}}.So this verifies our answer