X^2+11x+30>0 please solve

Solve the following inequality. Write the answer in interval notation

(x + 5)(x + 6) so the zeroes are –5 and –6. Since the parabola opens upward, it is negative between the two zeroes, so the quadratic is greater than 0 for x < –6 and x > –5.
In interval notation, (–∞, –6) U (–5, ∞)

The given expression is:
x^2+11x+30>0
First of all,apply the factorization method to get the factor of the equation,
as:
x^2+11x+30>0
x^2+6x+5x+30>0
Taking common:
x(x+6)+5(x+6)>0
(x+5)(x+6)>0
Now,equate the factors
either; , or;
x+5>0 , x+6>0
x>-5 , x>-6
So,
the required values are: -5 & -6
That's all.

There are many good ways to do this, one of which is to graph y= x^2+11x+30, then see where it is positive.

I would factor and make a line test.

(x+5)(x+6)>0

First find the zeroes: -5, and -6. Put them on a number line:

--------(-6)-------(-5)------- now test each part of the line. Pick any number in each of the three parts. plug into the factored expression. You only care about the sign.

-10:( + ) -5.5: ( - ). 10.( +)

Since we wanted positive only, x<-6 or > -5

Interval notation ( - infinity, -6) U ( -5, infinity)

Hoping this helps!

x^2+11x+30>0

x² + 11x > -30
x² + 11x + 121/4 > 121/4 - 30
(x+11/2)² > 1/4
x+11/2 > 1/2 => x > -5
x+11/2 < -1/2 => x < -6

x = all real numbers except [-6, -5]
(if you need it in some other notation, help yourself)

factor into (x+5)(x+6)>0

leaves two inequalities:

(x+5)>0 and (x+6)>0

the first leads to x >-5 and the second is x > -6 , so the answer is (-6, Infinity)

(x+6)(x+5)>0
so x=-6 x=-5