Question 144457
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 {{{2*x^2+8*x+1=0}}} ( notice {{{a=2}}}, {{{b=8}}}, and {{{c=1}}})





{{{x = (-8 +- sqrt( (8)^2-4*2*1 ))/(2*2)}}} Plug in a=2, b=8, and c=1




{{{x = (-8 +- sqrt( 64-4*2*1 ))/(2*2)}}} Square 8 to get 64  




{{{x = (-8 +- sqrt( 64+-8 ))/(2*2)}}} Multiply {{{-4*1*2}}} to get {{{-8}}}




{{{x = (-8 +- sqrt( 56 ))/(2*2)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (-8 +- 2*sqrt(14))/(2*2)}}} 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 = (-8 +- 2*sqrt(14))/4}}} Multiply 2 and 2 to get 4


So now the expression breaks down into two parts


{{{x = (-8 + 2*sqrt(14))/4}}} or {{{x = (-8 - 2*sqrt(14))/4}}}



Now break up the fraction



{{{x=-8/4+2*sqrt(14)/4}}} or {{{x=-8/4-2*sqrt(14)/4}}}



Simplify



{{{x=-2+sqrt(14)/2}}} or {{{x=-2-sqrt(14)/2}}}




So our answers are



{{{x=-2+sqrt(14)/2}}} or {{{x=-2-sqrt(14)/2}}}



which approximate to 



{{{x=-0.129171306613029}}} or {{{x=-3.87082869338697}}}