Question 407835


{{{3x^2-7x-13=0}}} Start with the given equation.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=3}}}, {{{b=-7}}}, and {{{c=-13}}}



Let's use the quadratic formula to solve for x



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



{{{x = (-(-7) +- sqrt( (-7)^2-4(3)(-13) ))/(2(3))}}} Plug in  {{{a=3}}}, {{{b=-7}}}, and {{{c=-13}}}



{{{x = (7 +- sqrt( (-7)^2-4(3)(-13) ))/(2(3))}}} Negate {{{-7}}} to get {{{7}}}. 



{{{x = (7 +- sqrt( 49-4(3)(-13) ))/(2(3))}}} Square {{{-7}}} to get {{{49}}}. 



{{{x = (7 +- sqrt( 49--156 ))/(2(3))}}} Multiply {{{4(3)(-13)}}} to get {{{-156}}}



{{{x = (7 +- sqrt( 49+156 ))/(2(3))}}} Rewrite {{{sqrt(49--156)}}} as {{{sqrt(49+156)}}}



{{{x = (7 +- sqrt( 205 ))/(2(3))}}} Add {{{49}}} to {{{156}}} to get {{{205}}}



{{{x = (7 +- sqrt( 205 ))/(6)}}} Multiply {{{2}}} and {{{3}}} to get {{{6}}}. 



{{{x = (7+sqrt(205))/(6)}}} or {{{x = (7-sqrt(205))/(6)}}} Break up the expression.  



So the answers are {{{x = (7+sqrt(205))/(6)}}} or {{{x = (7-sqrt(205))/(6)}}} 



If you need more help, email me at <a href="mailto:jim_thompson5910@hotmail.com">jim_thompson5910@hotmail.com</a>


Also, please consider visiting my website: <a href="http://www.freewebs.com/jimthompson5910/home.html">http://www.freewebs.com/jimthompson5910/home.html</a> and making a donation. Thank you


Jim