Question 475179
Yes it is possible.

{{{ ax^2 + bx + c - 7 = 0}}}
{{{a(x^2 + (b/a)x) + c - 7 = 0}}}

{{{a(x^2 + (b/a)x + b^2/(4a^2)) + c - 7 = b^2/(4a)}}}

{{{a(x + b/(2a))^2  + c - 7 - b^2/(4a) = 0}}}

{{{(x + b/(2a))^2  + (4ac - 28a - b^2)/(4a^2) = 0}}}
{{{(x + b/(2a))^2  - (b^2 - 4ac + 28a )/(4a^2) = 0}}}

{{{( x + b/(2a) - sqrt((b^2 - 4ac + 28a )/(4a^2)))( x + b/(2a) + sqrt((b^2 - 4ac + 28a )/(4a^2))) = 0}}}

It's routine from here...