Finding a real root in a cubic equation? I am trying to find a real root of the following cubic.
$x^3-6x^2+14x-15=0$
I did $x=y-\frac{b}{3a}$
which is $x=y+2$
I plugged my substitution to get the depressing cubic.
$(y+2)^3-6(y+2)^2+14(y+2)-15=0$
which is 
$y^3+2y=3$
then I set up the method for solving.
$y=s-t$ , first
$(3st=2)$,    second
$s^3-t^3=3$ 
So then for the second equation I got $s=\frac{2}{3t}$
Plugged it into my third and simplified 
$t^6+3t^3-\frac{8}{27}$
then Use the qudratic set $u=t^3$
so then I get
$u^2+3u-\frac{8}{27}$
Using the quadratic I got $-\frac{3}{2}+\frac{\sqrt{59/27}}{2}$
as my u which is t^3
take the cube root and I got $t=\sqrt[3]{\frac{-3}{2}+\frac{\sqrt{59/27}}{2}}$
so the s is $s=\sqrt[3]{\frac{3}{2}+\frac{\sqrt{59/27}}{2}}$
and since y=s-t you can plug than back into my $x=y+2$ equation to find x.
But I am not sure if I did this right

Note that $x=3$ is a solution. We used the Rational Roots Theorem, since cubics in exercises often have a rational root. Exercises and real life tend to differ.
When the depressed cubic was reached (correctly) the root $y=1$ was available by inspection.
