Question 604536
{{{ x^3+6x=10 }}}
It's a cubic equation. It has no x^2 term but the x^3 makes it a cubic.
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{{{ x^3+6x-10 = 0}}}
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x = {q + [q^2 + (r-p^2)^3]^1/2}^1/3 + {q - [q^2 + (r-p^2)^3]^1/2}^1/3 + p
where 

p = -b/(3a),   q = p^3 + (bc-3ad)/(6a^2),   r = c/(3a)
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a = 1, b = 0, c = 6, d = -10
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p = 0
q = 5
r = 2
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x = {5 + [25 + 8]^1/2}^1/3 + {5 - [25 + 8]^1/2}^1/3
= {{{root(3,(5 + sqrt(33))) + root(3,(5 - sqrt(33)))}}}
x =~ 1.300270977 the only real number solution