Question 764114
if you have equation like this:

{{{x-2=sqrt(x+10)}}} ...to solve for {{{x}}} you need square both sides first in order to get rid of square root

{{{(x-2)^2=(sqrt(x+10))^2}}}......now we can cancel root sign and power of {{{2}}} on right side

{{{(x-2)^2=(x+10)}}}

{{{x^2-4x+4=x+10}}}...now add {{{-x-10}}} to both sides

{{{x^2-4x+4-x-10=cross(x)+cross(10)-cross(x)-cross(10)}}}

{{{x^2-4x+4-x-10=0}}}

{{{x^2-5x-6=0}}}....now, we can use quadratic formula to solve for {{{x}}}

{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} ...as you can see, {{{a=1}}}, {{{b=-5}}}, and {{{c=-6}}}

{{{x = (-(-5) +- sqrt((-5)^2-4*1*(-6) ))/(2*1) }}}

{{{x = (5 +- sqrt(25+24 ))/2 }}}

{{{x = (5 +- sqrt(49 ))/2 }}}

{{{x = (5 +- 7)/2 }}}

solutions:

{{{x = (5 + 7)/2 }}}

{{{x = 12/2 }}}

{{{x = 6 }}}

or

{{{x = (5- 7)/2 }}}

{{{x = -2/2 }}}

{{{x = -1 }}}

this means {{{x-2}}} is equal to {{{sqrt(x+10)}}} if {{{x=6}}} and {{{x=-1}}}, and graphs intersecting at these points

now, we can see it on a graph:

 {{{ graph( 600, 600, -10, 10, -10, 10, x-2, sqrt(x+10),-sqrt(x+10)) }}}