Question

`3(x+2)2+4(x+2)+1=0\\`

Answer 1

Explanation:

`\Large \boldsymbol {} Let : t=x+2 \ \ ; \ \ t^2=(x+2)^2\\\\3(x+2)^2+4(x+2)+1 =0 \\\\\\3t^2+4t+1=0 \\\\D=16-12=4 \\\\ t_(1)=(-4-2)/(6) =-1 \ \ ; \ \ \boxed{x_1=-3 }\\\\\\ t_2=(-4+2)/(6) =-(1)/(3) \ \ ; \ \boxed{x_2=-2(1)/(3) }`

Answer 2

-3

-7/3

Explanation:

Let u=x+2

We need to now solve 3u^2+4u+1=0 for u.

I'm going to try factoring. 3u (u)=3u^2 and 1(1)=1 so I'm going to try factored form (3u+1)(u+1).

Since 3u(1)+1(u)=4u, we are done.

Since (3u+1)(u+1)=0, then either 3u+1=0 or u+1=0.

Let's solve u+1=0 by subtracting 1 on both sides which obtains for us that u=-1.

Let's solve 3u+1=0 by subtracting 1 and then dividing both sides by 3 which obtains for us u=-1/3.

Since u=x+2, we are down to just solving the following:

x+2=-1 and x+2=-1/3

Both equations require us to subtract 2 on both sides.

The solutions are

x=-1-2=-3 and x=-1/3-2=-7/3