How to solve equations to the fourth power?

One method is to set $x=y^2$ and rearrange this as a quadratic equation $$x^2-153x+1296=0$$

Here's the solution:

$$y^4-153y^2 +1296 = 0$$

$$ y^4 -144y^2-9y^2+1296 = 0$$

$$ y^2(y^2-144) -9(y^2-144) = 0$$

$$ (y^2-9)(y^2-144)=0$$ $$ (y^2-3^2)(y^2-12^2)=0$$

note that $a^2 - b^2 = (a-b)(a+b)$

Can you see how now? I trust you can finish the rest

While the convert-to-quadratic approach works here, and other answers have provided this approach, it's worth noting that rational solutions to quartics with integer coefficients work similarly to quadratics.

That is, if you have a quartic of the form $$ ax^4+bx^3+cx^2+dx+e=0 $$ where all coefficients are integers, then all rational solutions must be of the form $$ x = \frac{\text{divisor of }e}{\text{divisor of }a} $$ In this case, we have $a=1$, $c=-153$, $e=1296$, and $b=d=0$. So we seek a solution of the form $$ x = \frac{\text{divisor of }1296}{\text{divisor of }1} $$ As the only divisors of 1 are 1 and -1, we need only look at integer values. 1296 has the prime factorisation $$ 1296 = 2^4\times 3^4 $$ and from here, we just need to start substituting trial values in until we find a root. With $P(x)=x^4-153x^2+1296$, noting that only even powers of $x$ appear (and thus we need not consider the negatives separately), we have $$ P(1) = 1 - 153 + 1296 = 1144 \neq 0\\ P(2) = 16 - 612 + 1296 = 700 \neq 0\\ P(3) = 81 - 1377 + 1296 = 0 $$ and thus $x=3$ is a solution. Again noting that only even powers appear, we see that $x=-3$ is also a solution. It's fairly trivial to continue in this way to find the other two roots, as well.

Note that this method won't find irrational roots, however.