Solving $x^4-15x^2-10x+ 24 = 0$ using Ferrari’s method 
Ferrari’s method for solving a quartic equation
$$x^4-15x^2-10x+ 24 = 0$$
begins by writing:$$x^4= 15x^2+ 10x-24$$and then adding a term of
  the form:$$-2bx^2+b^2$$to both sides.
(a) Explain why this is good idea and what it accomplishes.
(b) Use $b= 7$ to find the two quadratic equations that yield the
  solutions.

$(a)$
I have
$$\begin{align*}x^4-2bx^2+b^2&=15x^2+ 10x-24-2bx^2+b^2\\\\
&=(15-2b)x^2+ 10x+(b^2-24)
\end{align*}$$
I then notice that $$x^4-2bx^2+b^2=(x^2-b)^2$$
so we get
$$(x^2-b)^2=(15-2b)x^2+ 10x+(b^2-24)$$
Pluggin in $b=7$ I get
$$(x^2-7)^2=x^2+ 10x+25$$
but I fail to see why this helps with regards to solving for $x$.
Any hints to lean me in the direction would be much appreciated.

$$(x^2-7)^2=x^2+ 10x+25\implies (x^2-7)^2-(x+5)^2=0$$
$$(x^2-x-12)(x^2+x-2)=0$$
Now solve each quadratic equation for $ x$ 
