Question 323992
Substitute, let {{{u=5+sqrt(x)}}}
{{{u^2-12u+33=0}}}
Use the quadratic formula,
{{{u = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
{{{u = (12 +- sqrt( 12^2-4*1*(33) ))/(2*1) }}} 
{{{u = (12 +- sqrt( 144-132 ))/2 }}} 
{{{u = (12 +- sqrt( 12 ))/2 }}}
{{{u = (12 +- 2sqrt( 3 ))/2 }}}
{{{u = 6 +- sqrt( 3 ) }}}
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{{{5+sqrt(x)=6 +- sqrt(3)}}}
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{{{5+sqrt(x)=6+sqrt(3)}}}
{{{sqrt(x)=1+sqrt(3)}}}
{{{x=(1+sqrt(3))^2}}}
{{{x=1+2sqrt(3)+3}}}
{{{highlight(x=4+2sqrt(3))}}} or approximately,
{{{x=7.73}}}
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{{{5+sqrt(x)=6 - sqrt(3)}}}
{{{sqrt(x)=1-sqrt(3)}}}
{{{sqrt(x)<0}}}
No solution.
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{{{graph(300,300,-5,10,-2,2,(5+sqrt(x))^2-12(5+sqrt(x))+33)}}}