Solve $(x^2 + 5)^2 - 15(x^2 + 5) + 54 = 0$ I got the square root of 14 and 11 but the answer book states that these answers are wrong. Can someone help me? I used this formula to find the individual roots 
$x = -\frac{p}{2} \pm \sqrt{(\frac{p}{2})^2 - q}$ 
 A: Let $a = (x^2 + 5)$. Then $$(a-9)(a-6)=0$$
$$\implies a = 9\; or \;a = 6$$
$$\implies x^2+5 = 9 \; or \;6$$
$$\implies x = \pm\sqrt4\; or \pm\sqrt1$$
$$\implies x = \pm2\; or \pm1$$
This kind of equation is called a biquadratic equation. Cheers!
A: Using middle term factor,  $$(x^2+5)^2-(9+6)(x^2+5)+6\cdot9=0$$
$$\implies (x^2+5)(x^2+5-9)-6(x^2+5-9)=0$$
$$\implies (x^2-4)(x^2-1)=0$$
$\implies x^2-1=0$ or $x^2-4=0$
Alternatively, using quadratic equation formula  for $x^2+5=\frac{15\pm\sqrt{15^2-4\cdot1\cdot 54}}{2\cdot1}=\frac{15\pm 3}2=9$ or $6$
A: Given: 
$(x^2 + 5)^2 - 15(x^2 + 5) + 54 = 0$
Let a = $(x^2+5)$ which gives us a simple quadratic trinomial of the form $ax^2+bx+c=0.$
$a^2-15a+54=0$
Factoring this we get:
$(a-9)(a-6)=0$
$a-9=0$ 
$a-6=0$
$\implies a=9$ or $a=6$
$x^2+5=9$
$x^2+5=6$
$\implies x = \pm2$ or $\pm 1$
You can check that these solutions are correct by simply plugging values back in to see if you get $0$.
$a^2-15a+54=0 \implies$ $(9)^2 - 15(9) + 54 = 0 \implies 81-135+54=0$ 
