Question 2855
Compare with standard form
{{{ax^2+bx+c=0}}}
We get,
a=1,b=8,c=14.
Formula for Quadratic Equations is:


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


Then substituting the values of a,b and c,we get:


{{{x = (-8 +- sqrt( 8^2-4*1*14 ))/(2*1) }}} 

Then step by step we get,


{{{x = (-8 +- sqrt( 64-4(1)(14) ))/2 }}}


{{{x = (-8 +- sqrt( 64-56 ))/2 }}}


{{{x = (-8 +- sqrt( 8 ))/2 }}}

Therefore we have 2 values possible for x,which are:

{{{x = (-8 + sqrt( 8 ))/2 }}} OR {{{x = (-8 - sqrt( 8 ))/2 }}}


Note that square root of 8 is: 2.828

On further solving as per formula,we get possible values of x as:

(-8+2.828)/2 OR (-8-2.828)/2

Resulting in rounded off values of x as:

-2.586 OR -5.414

Therefore,
Solution Set={-2.586,-5.414}

Hope you followed this,
Best of luck.