Question

find the real solutions of the equation. 4(x+1)^2+14(x+1)+6=0

Answer 1

let t = x+1
then 4(x+1)^2+14(x+1)+6=0 is just 4t^2 + 14t + 6 = 0
note that 4*6 = 24 = 2*12 and 2+12 = 14, thus
4t^2 + 14t + 6 = 0 can be factored (by grouping)
4t^2 + 2t + 12t + 6 = 0
2t(2t + 1) + 6(2t + 1) = 0
(2t + 1)(2t + 6) = 0
so 2t + 1 = 0 which means that t = -1/2
and 2t+6 = 0 which means that t = -3

Now since t = x+1 we have that
x + 1 = -1/2 which means that x = -3/2 and
x + 1 = -3 which means that x = -4

To check:
4(-3/2+1)^2+14(-3/2+1)+6
4(-1/2)^2 + 14(-1/2) + 6
4(1/4) -7 + 6
1-7+6
0 check :)

4(-4+1)^2+14(-4+1)+6
4(-3)^2 + 14(-3) + 6
4(9) -42 + 6
36 - 42 + 6
0 check :)

So the two solutions are x = -3/2 and x = -4

Answer 2

Substitute (x+1) by a for example. The equation becomes \[4a ^{2}+14a+6=0\]