Question 102474
{{{x^2-13x=-10}}} Start with the given equation



{{{x^2-13x+10=0}}} Add 10 to both sides



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



Starting with the general quadratic


{{{ax^2+bx+c=0}}}


the general solution using the quadratic equation is:


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




So lets solve {{{x^2-13*x+10=0}}} ( notice {{{a=1}}}, {{{b=-13}}}, and {{{c=10}}})





{{{x = (--13 +- sqrt( (-13)^2-4*1*10 ))/(2*1)}}} Plug in a=1, b=-13, and c=10




{{{x = (13 +- sqrt( (-13)^2-4*1*10 ))/(2*1)}}} Negate -13 to get 13




{{{x = (13 +- sqrt( 169-4*1*10 ))/(2*1)}}} Square -13 to get 169  (note: remember when you square -13, you must square the negative as well. This is because {{{(-13)^2=-13*-13=169}}}.)




{{{x = (13 +- sqrt( 169+-40 ))/(2*1)}}} Multiply {{{-4*10*1}}} to get {{{-40}}}




{{{x = (13 +- sqrt( 129 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (13 +- sqrt(129))/(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>)




{{{x = (13 +- sqrt(129))/2}}} Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


{{{x = (13 + sqrt(129))/2}}} or {{{x = (13 - sqrt(129))/2}}}



Now break up the fraction



{{{x=+13/2+sqrt(129)/2}}} or {{{x=+13/2-sqrt(129)/2}}}



Simplify



{{{x=13 / 2+sqrt(129)/2}}} or {{{x=13 / 2-sqrt(129)/2}}}



So these expressions approximate to


{{{x=12.1789083458003}}} or {{{x=0.821091654199726}}}



So our solutions are:

{{{x=12.1789083458003}}} or {{{x=0.821091654199726}}}


Notice when we graph {{{x^2-13*x+10}}}, we get:


{{{ graph( 500, 500, -9.17890834580027, 22.1789083458003, -9.17890834580027, 22.1789083458003,1*x^2+-13*x+10) }}}


when we use the root finder feature on a calculator, we find that {{{x=12.1789083458003}}} and {{{x=0.821091654199726}}}.So this verifies our answer