Question 130755
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^2-5*x+3=0}}} ( notice {{{a=1}}}, {{{b=-5}}}, and {{{c=3}}})





{{{x = (--5 +- sqrt( (-5)^2-4*1*3 ))/(2*1)}}} Plug in a=1, b=-5, and c=3




{{{x = (5 +- sqrt( (-5)^2-4*1*3 ))/(2*1)}}} Negate -5 to get 5




{{{x = (5 +- sqrt( 25-4*1*3 ))/(2*1)}}} Square -5 to get 25  (note: remember when you square -5, you must square the negative as well. This is because {{{(-5)^2=-5*-5=25}}}.)




{{{x = (5 +- sqrt( 25+-12 ))/(2*1)}}} Multiply {{{-4*3*1}}} to get {{{-12}}}




{{{x = (5 +- sqrt( 13 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)





{{{x = (5 +- sqrt(13))/2}}} Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


{{{x = (5 + sqrt(13))/2}}} or {{{x = (5 - sqrt(13))/2}}}




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So the exact solutions are



{{{x = (5 + sqrt(13))/2}}} or {{{x = (5 - sqrt(13))/2}}}




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When you use a calculator, you will find that the approximate solutions are



{{{x=4.30277563773199}}} or {{{x=0.697224362268005}}}