Mathematics induction on inequality: $2^n \ge 3n^2 +5$ for $n\ge8$ I want to prove

$2^n \ge 3n^2 +5$--call this statement $S(n)$--for $n\ge8$

Basis step with $n = 8$, which $\text{LHS} \ge \text{RHS}$, and $S(8)$ is true. Then I proceed to inductive step by assuming $S(k)$ is true and so

$2^k \ge 3k^2 +5 $

Then $S(k+1)$ is

$2^{k+1} \ge 3(k+1)^2 + 5$

I need to prove it so I continue by

$2^{k+1} \ge 3k^2 +5$
$2(2^k) \ge 2(3k^2 +5)$

I'm stuck here...don't know how to continue, please explain to me step by step. I search for whole day already, all give answer without explanation. I can't understand, sorry for the trouble.

Observe that since $k\ge8>6$, then $$3(k+1)^2+5=3k^2+6k+8<3k^2+k^2+8=4k^2+8<6k^2+8<6k^2+10.$$
Rewriting the far right expression as $2(3k^2+5)$, we use the fact that $S(k)$ holds (and the fact that multiplication by $2$ preserves order) to conclude that $$3(k+1)^2+5<2(3k^2+5)\leq 2(2^k)=2^{k+1},$$ so $S(k+1)$ holds.
