<s>[INST] 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 
$x = -\frac{p}{2} \pm \sqrt{(\frac{p}{2})^2 - q}$ 
 [/INST] 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 $0$.
$a^2-15a+54=0 \implies$ $(9)^2 - 15(9) + 54 = 0 \implies 81-135+54=0$ 
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