<PRE><B>Question 212144</B>
Solve for x in the following equation


{{{sqrt(2x+5)=x-5}}}


Step 1.  Square Both Sides of Equation


{{{(sqrt(2x+5))^2=(x-5)^2}}}


{{{2x+5=x^2-10x+25}}}


Step 2.  Simplify to get a quadratic equation.  Subtract 2x+5 from both sides of equation


{{{2x+5-(2x+5)=(x-5)^2=x^2-10x+25-(2x+5)}}}


{{{0=x^2-10x+25-(2x+5)}}}


{{{0=x^2-8x+20}}}


Step 3.  Now use the quadratic equation


{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}


where a=1, b=-8 and c=20 and follow steps below:


*[invoke quadratic &quot;x&quot;, 1, -2, -3 ]


Step 4.  x=3 and x=-1 are solutions to the equation.


For Step-By-Step Videos on Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra.</PRE>