Unsure how to solve equation I have the problem. Solve: $x^3 - 3x^2 + 2x - 5 = 0$. I've done similar, simpler ones like this, but in this case I don't know. Can someone at least tell me what you'd even call this exact type of problem so I can google it and learn how to do this and work on it more?
Here's a way to find all complex $x$ satisfying
$$x^3-3x^2+2x-5=0$$
First we want to remove the $-3x^2$ term, this can be done by doing the substitution $x=y+1$
$$(y+1)^3-3(y+1)^2+2(y+1)-5=0$$
Expanding this yields
$$y^3-y-5=0$$
Substitute $y=\frac\lambda z+z$, this also means that $z = \frac12\left(y+\sqrt{y^2-4\lambda}\right)$

$$-5-z-\frac\lambda z+\left(z+\frac\lambda z\right)^3=0$$

Multiply both sides with $z^3$ and collect in terms of $z$

$$z^6+z^4(3\lambda-1)-5z^3+z^2\left(3\lambda^2-\lambda\right)+\lambda^3=0$$

Substitute $\lambda=\frac13$ and then $u=z^3$

$$u^2-5u+\frac1{27}=0$$

Find the solution where $\pm$ is a $+$

$$u=\frac1{18}\left(45+\sqrt{2013}\right)$$

Substitute back $u=z^3$

$$z^3=\frac1{18}\left(45+\sqrt{2013}\right)$$

The cube root of that is $\frac{\sqrt[3]{45+\sqrt{2013}}}{\sqrt[3]{2}~3^{2/3}}$

Multiply that with the third roots of unity.

$$z=\frac{\sqrt[3]{\frac12\left(45+\sqrt{2013}\right)}}{3^{2/3}}$$
$$z=-\frac{\sqrt[3]{\frac12\left(-45-\sqrt{2013}\right)}}{3^{2/3}}$$
$$z=\frac{(-1)^{2/3}\sqrt[3]{\frac12\left(45+\sqrt{2013}\right)}}{3^{2/3}}$$

Substituting back for $z = \frac y2+\frac12\sqrt{y^2-\frac43}$

$$\frac y2+\frac12\sqrt{y^2-\frac43}=\frac{\sqrt[3]{\frac12\left(45+\sqrt{2013}\right)}}{3^{2/3}}$$
$$\frac y2+\frac12\sqrt{y^2-\frac43}=-\frac{\sqrt[3]{\frac12\left(-45-\sqrt{2013}\right)}}{3^{2/3}}$$
$$\frac y2+\frac12\sqrt{y^2-\frac43}=\frac{(-1)^{2/3}\sqrt[3]{\frac12\left(45+\sqrt{2013}\right)}}{3^{2/3}}$$

Solve the radical equation for $y$

$$y=\sqrt[3]{\frac2{3\left(45+\sqrt{2013}\right)}}+\frac{\sqrt[3]{\frac{45+\sqrt{2013}}{2}}}{3^{2/3}}$$
$$y=(-1)^{2/3}\sqrt[3]{\frac2{3\left(45+\sqrt{2013}\right)}}-\frac{\sqrt[3]{\frac{-45-\sqrt{2013}}{2}}}{3^{2/3}}$$
$$y=\frac{(-1)^{2/3}\sqrt[3]{\frac{45+\sqrt{2013}}{2}}}{3^{2/3}}-\sqrt[3]{\frac{-2}{3\left(45+\sqrt{2013}\right)}}$$

Substitute back $y=x-1$ and add $1$ to both sides

$$x=1+\sqrt[3]{\frac2{3\left(45+\sqrt{2013}\right)}}+\frac{\sqrt[3]{\frac{45+\sqrt{2013}}{2}}}{3^{2/3}}$$
$$x=1-(-1)^{2/3}\sqrt[3]{\frac2{3\left(45+\sqrt{2013}\right)}}-\frac{\sqrt[3]{\frac{-45-\sqrt{2013}}{2}}}{3^{2/3}}\approx$$
$$x=1+\frac{(-1)^{2/3}\sqrt[3]{\frac{45+\sqrt{2013}}{2}}}{3^{2/3}}-\sqrt[3]{\frac{-2}{3\left(45+\sqrt{2013}\right)}}$$

Here's some approximate versions of these numbers

$$
\begin{array}{rclcl}
x&amp;\approx&amp;2.90416085913492\\
x&amp;\approx&amp;0.04791957043254&amp;-&amp;1.311248044077123i\\
x&amp;\approx&amp;0.04791957043254&amp;+&amp;1.311248044077123i\\
\end{array}
$$
This equation has a single real root. The solution has a closed form formula, but you can also find the numerical value by Newton's iterations
$$x_{n+1}=x_n-\frac{x_n^3-3x_n^2+2x_n-5}{3x_n^2-6x_n+2},$$ starting from, say, $x_0=0$. (The denominator is the derivative of the numerator.)
UPDATE:
After finding the single real root, you can show that there is no other by deflation (divide by $x-r$ and show that the quadratic quotient has no real root).