Question 353789
{{{x^2 + 7x - 30 = 0}}}
Add {{{30}}} to both sides
{{{x^2 + 7x = 30}}}
Take 1/2 of {{{7}}}, square it, and 
add it to both sides
{{{x^2 + 7x + (7/2)^2 = 30 + (7/2)^2}}}
{{{x^2 + 7x + 49/4 = 30 + 49/4}}}
{{{x^2 + 7x + 49/4 = 120/4 + 49/4}}}
{{{x^2 + 7x + 49/4 = 169/4}}}
{{{(x + 7/2)^2 = (13/2)^2}}}
Both sides are now perfect squares. The
right side doesn't have to be a perfect
square, but the solution is neater if it is.
Now take the square root of both sides
{{{x + 7/2 = 13/2}}}
{{{x = 13/2 - 7/2}}}
{{{x = 6/2}}}
{{{x = 3}}} 
Taking the (-) square root:
{{{x + 7/2 = -13/2}}}
{{{x = -13/2 - 7/2}}}
{{{x = -20/2}}}
{{{x = -10}}}
From these I get {{{x - 3 = 0}}}
and {{{x + 10 = 0}}}
{{{(x - 3)*(x + 10) = x^2 + 7x - 30}}}