Question 899481
Solve by any method:

8x^2= -22x+6
<pre>
{{{8x^2 = - 22x + 6}}}
{{{8x^2 + 22x - 6 = 0}}}
{{{2(4x^2 + 11x - 3) = 2(0)}}}
{{{4x^2 + 11x - 3 = 0}}}
{{{4x^2 + 12x - x - 3 = 0}}} ------ Replacing "+11x" with "+12x - x"
{{{4x(x + 3) - 1(x + 3) = 0}}}
{{{4x - 1)(x + 3) = 0}}}
4x - 1 = 0            OR          x + 3 = 0
    4x = 1            OR              x = 0 - 3
     x = {{{highlight_green(1/4)}}}          OR              {{{highlight_green(x = - 3)}}}
You can do the check!! 

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