Question 149309
By solve, I assume you mean find the zeros where f(x)=0.
{{{f(x)=x^3-4x^2+7x-2}}}
Typically the easiest way is to graph and look for the function crossing the x axis.
{{{ graph( 300, 300, -20, 20
, -10, 10, x^3-4x^2+7x-2) }}}
Looks like it crosses once at about x=1/3.
Since there are three roots for a cubic equation, the other two roots are complex.
Using Excel to window down the real solution gives x=0.349370808.
That gives a value of f(x)=-2.6e-9. That's pretty close to zero.
The other two roots are complex.
{{{x=1.82531 +- 1.5468689i}}}
There is an explanation of how to find zeros for a cubic equation at http://www.1728.com/cubic.htm.

{{{ graph( 300, 300, -1, 2
, -10, 10, x^3-4x^2+7x-2) }}}
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The other function is simpler.
{{{f(x)=(4x-7)^20}}}
This equals zero when {{{(4x-7)=0}}}
{{{4x-7=0}}}
{{{x=7/4}}}
There are 20 roots at x=7/4.
{{{ graph( 300, 300, -2, 3, -10, 10, (4x-7)^20) }}}