Question 412980
I can help
{{{6x^2 + 5x = 6}}}
first, divide through by {{{6}}}, so 
{{{x^2}}} has coefficient = {{{1}}}
{{{x^2 + (5/6)*x = 1}}}
Now take 1/2 of the coefficient of {{{x}}},
square it, and add it to both sides
{{{ x^2 + (5/6)*x + (5/12)^2 = 1  +  (5/12)^2}}}
{{{x^2 +  (5/6)*x + 25/144 = 144/144 + 25/144}}}
{{{ x^2 +  (5/6)*x + 25/144 = 169/144 }}}
Both sides are now a perfect square
{{{ (x + 5/12)^2 = (13/12)^2}}}
Take the square root of both sides
{{{x + 5/12 = 13/12}}}
{{{x = 8/12}}}
{{{x = 2/3}}}
And there is also a negative square root
{{{x + 5/12 = -13/12}}}
{{{x = -3/2}}}
The roots are 2/3 and -3/2
I can check this by
{{{(x - 2/3)*(x + 3/2) = 0}}}
{{{x^2 -( 2/3)*x + (3/2)*x - 1 = 0}}}
{{{x^2 -(4/6)*x + (9/6)*x = 1}}}
Multiply through by {{{6}}}
{{{6x^2 - 4x + 9x = 6}}}
{{{6x^2 + 5x = 6}}}
OK