Question 548975
{{{7x^2+10x=-2}}}

We will first put everything on one side of the equation (as in most classes, students are most familiar with completely the square when it is in the form ax^2+bx+c

this gives us:
{{{7x^2+10x+2=0}}}
Now we need to make the constant in front of x^2 1, so we will factor out a 7 from our first two terms:
{{{7(x^2+10/7*x)+2=0}}}
Now the rules for completing the square say we must take the constant in front of the x and square one half of this number and add and subtract this number.  

So our constant is 10/7.  One half of 10/7 is 
10/7*1/2=10/14
and this squared is 
(10/14)^2=(5/7)^2=25/49

So we will add and subtract this number (so essentially we are adding 0 to our equation, which does not change it). 
{{{7(x^2+10/7*x+25/49-25/49)+2=0}}}
We will now take our -25/49 out of the bracket (remembering to multiply it by 7 when we take it out).  this leaves:
{{{7(x^2+10/7*x+25/49)+2-25/49*7=0}}}

Almost there.... 
We will complete the square inside of the brackets.  Essentially we just write it in the form (x+a)^2 where a is 1/2 of the constant in front of the x term originally (we found this was 10/14=5/7

So we have:
{{{7(x+5/7)^2+2-25/7=0}}}
{{{7(x+5/7)^2+14/7-25/7=0}}}
{{{7(x+5/7)^2-11/7=0}}}


Now it is just solving for x, so let's isolate x:
{{{7(x+5/7)^2-11/7=0}}}
{{{7(x+5/7)^2=11/7}}}
{{{7(x+5/7)^2/7=(11/7)/7}}}
{{{(x+5/7)^2=11/49}}}
{{{(x+5/7)=+-sqrt(11/49)}}}
{{{x=+-sqrt(11/49)-5/7}}}


This can be simplified on your calculator... but I will leave that to you.  
Hopefully this helps... let me know if you are still unsure.