Question 1185300
<pre>

{{{sqrt(2x^2 + 5x + 6)}}}{{{""=""}}}{{{x}}}

We square both sides:

{{{(sqrt(2x^2 + 5x + 6))^2}}}{{{""=""}}}{{{(x)^2}}}

When we square a square root we take away the square and the square root:

{{{2x^2 + 5x + 6}}}{{{""=""}}}{{{x^2}}}

Subtract x<sup>2</sup> from both sides:

{{{x^2 + 5x + 6}}}{{{""=""}}}{{{0}}}

{{{(x+3)(x+2)}}}{{{""=""}}}{{{0}}}

Use the zero factor principle. Set each factor equal to 0:

x+3 = 0;   x+2 = 0
  x = -3;    x = -2

We substitute those in the original equation to see if both are
solutions to the original equations, or extraneous:

Substituting x = -3

{{{sqrt(2(-3)^2 + 5(-3) + 6)}}}{{{""=""}}}{{{-3}}}

{{{sqrt(2(9) - 15 + 6)}}}{{{""=""}}}{{{-3}}}

{{{sqrt(18 - 15 + 6)}}}{{{""=""}}}{{{-3}}}

{{{sqrt(9)}}}{{{""=""}}}{{{-3}}}

{{{3}}}{{{""=""}}}{{{-3}}}

That's false.  So -3 is an extraneous solution, so we discard it.

Substituting -2

{{{sqrt(2(-2)^2 + 5(-2) + 6)}}}{{{""=""}}}{{{-2}}}

{{{sqrt(2(4) - 10 + 6)}}}{{{""=""}}}{{{-2}}}

{{{sqrt(8 - 10 + 6)}}}{{{""=""}}}{{{-2}}}

{{{sqrt(4)}}}{{{""=""}}}{{{-2}}}

{{{2}}}{{{""=""}}}{{{-2}}}

That's false.  So -2 is also an extraneous solution, so we discard it.

The equation has no solution.

Note: The symbol "√(" always means the POSITIVE square root. To indicate
the negative square root, there must be a negative sign before it "-√(".

Edwin</pre>