When solving a quadratic inequality, you solve the corresponding equation first, to calculate the point(s) of intersection. Just like with linear inequalities.
Then you decide on the solution using a sketch or the number line.

You have to know the meaning of the different signs being used.
If necessary, check domain and range for theory about intervals.

Example 1
a2 + 30 < –11a + 2
a2 + 30 = –11a + 2
a2 + 11a + 28 = 0
(a + 4)(a + 7) = 0
a = –4 or a = –7

Continue with sketch Continue with number line
There are two points of intersection and the graph is a upward-opening parabola.

That gives you this sketch: Look to the original inequality:
a2 + 30 < –11a + 2
Parabola < line
That is in between the points of intersection.

Solution: –7 < a < –4
Calculate whether or not the inequality is
correct for numbers outside and in
between the solutions.

Outside the points of intersection:
Take for example a = –8.
 (–8)2 + 30 < –11 × –8 + 2 94 < 90 Inc orrect!

In between the two solutions:
Take for example a = –6.
 (–6)2 + 30 < –11 × –6 + 2 66 < 68 Co rrect! Because 'in between' is correct,
the solution is:
–7 < a < –4

Example 2

 –2x2 – 4x – 69 ≤ –28x – 5 –2x2 – 4x – 69 = –28x – 5 –2x2 + 24x – 64 = 0 x2 – 12x + 32 = 0 (x – 4)(x – 8) = 0 x = 4 of x = 8

Continue with sketch Continue with number line
There are two points of intersection and the graph is a downward-opening parabola.

That gives you this sketch: Look to the original inequality:
 –2x2 – 4x – 69 ≤ –28x – 5 parabool ≤ lijn
That is outside the points of intersection!

Solution: x ≤ 4 or x ≥ 8
Calculate whether or not the inequality is
correct for numbers outside and in
between the solutions.

Outside the points of intersection:
Take for example x = 2.
 –2 × 22 – 4 × 2 – 69 ≤ –28 × 2 – 5 –85 ≤ –61 Co rrect!

In between the two solutions:
Take for example x = 5.
 –2 × 52 – 4 × 5 – 69 ≤ –28 × 5 – 5 –139 ≤ –145 Inc orrect! Because 'outside' is correct,
the solution is:
x ≤ 4 or x ≥ 8

## Watch out with:

### No intersections (see sketch)

Parabola < line or parabola ≤ line
Solution: for every value of x

Parabola > line or parabola ≥ line
Solution: for no value of x ### One point of intersection/tangent* (see sketch)

*A tangent in this context is the point
where the graphs touch.

Parabola < line
Solution: x ≠ 1
(Every value of x except x = 1)

Parabola > line
Solution: for no value of x

Parabola ≤ line
Solution: for every value of x

Parabola ≥ line
Solution: x = 1 tangent at x = 1