Question 5225
Caroline,

Try using y instead of {{{sqrt(x)}}}. x is {{{sqrt(x)}}} squared, or {{{y^2}}}. So, your original equation becomes

{{{y^2+y-6 = 0}}}. I will use my pluggable quadratic solver now:


*[invoke quadratic "y", 1, 1, -6]

since y is sqrt(x), y cannot be negative, so solution -3 is invalid. You have y=2, y is {{{sqrt(x)}}}, so x is 4.