Question 697088


solve the following algebraically:

{{{x-3= sqrt(x^2+3x)}}}

my work:

{{{(x-3)^2= sqrt (x^2+3x)^2}}}....perfect, you did it right

{{{x^2-18x+9=x^2+3x}}}....perfect

 
Then i subtracted {{{x^2+3x}}} from both sides and got:

{{{-21x+9=0}}} and pulled out a {{{3}}} and got {{{3(7x+3)=0}}} ...just finish this and solve for {{{x}}}

{{{3(7x+3)=0}}} ...this product will be equal to zero if and only if 

{{{7x+3=0}}}; so,

{{{7x=-3}}}

{{{x=-3/7}}}....or as decimal {{{x=-0.43}}}