x^2 + 5x > 14

To solve the inequality, first we will have to move 14 to the left side of the inequality:

==> x^2+ 5x - 14 > 0

Now we can factor the left side:

==> (x+ 7) (x-2 ) >0

Now we notcie that we have a product of two functions.

In order for the product we be greater that 0, we have two possible cases:

1. Both terms should be positivs:

==> x+ 7 > 0 and x-2 > 0

==> x > -7 and x > 2

==> x belongs to the interva ( 2, inf)

2. Both terms should be negative:

==> x+ 7 < 0 and x-2 < 0

==> x < -7 and x < 2

==> x belongs to the interval ( -inf, -7)

Then x = ( -inf, -7) U ( 2, inf)

OR x = R - **[-**7,2]

To solve x^2+5x>14.

We add (5/2)^2 to both sides in order that the left side becomes a complete square and solve the inequality.

x^2+ (5/2)^2 >14+(5/2)^2.

(x+5/2)^2 > 81/4.

We take square root of both sides remembering that the LH could be greater than the right only if

x+5/2 > 9/4 , x+5/2 < -9/5

x> 9/2-5/2, or x= -9/2-5/2.

x = 4/2 = 2. Or x < -9/2 -5/2 = -14/2 = -7.

Therefore x > 2 , or x< -7 are the solutions.

Therefore

Tally : Put x> 2 in x^2+5x : 2^2+5*10 >14 = RHS.

Put x= -7 in x^2+5x : (-7)^2 + 5(-7) > 49-35 = 14 = RHS.