Question 926130


{{{4x^2+x-20=0}}} Start with the given equation.



Notice that the quadratic {{{4x^2+x-20}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=4}}}, {{{B=1}}}, and {{{C=-20}}}



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



{{{x = (-B +- sqrt( B^2-4AC ))/(2A)}}} Start with the quadratic formula



{{{x = (-(1) +- sqrt( (1)^2-4(4)(-20) ))/(2(4))}}} Plug in  {{{A=4}}}, {{{B=1}}}, and {{{C=-20}}}



{{{x = (-1 +- sqrt( 1-4(4)(-20) ))/(2(4))}}} Square {{{1}}} to get {{{1}}}. 



{{{x = (-1 +- sqrt( 1--320 ))/(2(4))}}} Multiply {{{4(4)(-20)}}} to get {{{-320}}}



{{{x = (-1 +- sqrt( 1+320 ))/(2(4))}}} Rewrite {{{sqrt(1--320)}}} as {{{sqrt(1+320)}}}



{{{x = (-1 +- sqrt( 321 ))/(2(4))}}} Add {{{1}}} to {{{320}}} to get {{{321}}}



{{{x = (-1 +- sqrt( 321 ))/(8)}}} Multiply {{{2}}} and {{{4}}} to get {{{8}}}. 



{{{x = (-1+sqrt(321))/(8)}}} or {{{x = (-1-sqrt(321))/(8)}}} Break up the expression.  



So the exact solutions are {{{x = (-1+sqrt(321))/(8)}}} or {{{x = (-1-sqrt(321))/(8)}}} 

------------------------------------------------------------------------------------------------------------------------


If you need more one-on-one help, email me at <a href="mailto:jim_thompson5910@hotmail.com?Subject=I%20Need%20Algebra%20Help">jim_thompson5910@hotmail.com</a>. You can ask me a few more questions for free, but afterwards, I would charge you ($2 a problem to have steps shown or $1 a problem for answer only).


Alternatively, 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. Any amount is greatly appreciated as it helps me a lot. This donation is to support free tutoring. Thank you.


Jim

------------------------------------------------------------------------------------------------------------------------