Solving $x^2+3x+2=0$ gives two discrete points. The inequality $x^2+3x+2>0$ asks a different question entirely: for which infinite range of $x$-values is this true? A common and costly error is changing the inequality sign to "=" and treating the critical values as the whole answer — they're only the boundaries of the answer.

Step 1 — Standard form with $a>0$: Rearrange so one side is 0, and multiply by $-1$ if needed to make the $x^2$ coefficient positive (remember: multiplying/dividing by a negative flips the inequality sign). This guarantees you'll always be sketching a U-shape, removing one whole layer of decision-making.

Step 2 — Find critical values: Solve the related equation $ax^2+bx+c=0$ (factorise, or use the quadratic formula if it doesn't factorise).

Step 3 — Sketch: A simple U-shaped sketch crossing the $x$-axis at your two critical values (correctly ordered left-to-right) is non-negotiable — it is the only reliable way to avoid Step 4's main pitfall.

Step 4 — Interpret: Read the solution directly off the sketch.

• $f(x)>0$ (above axis) → the two outer "arms" → "OR": $x<p$ or $x>q$
• $f(x)<0$ (below axis) → the single "trough" between roots → "AND": $p<x<q$

Writing $-5>x>3$ for an "OR" solution. This is mathematically impossible — no number is both less than $-5$ and greater than $3$. The word "or" is not optional; it's the only correct way to describe a disjoint region: $x<-5$ or $x>3$.

Use the quadratic formula for the critical values, same method otherwise.

Worked example: Find the integer values satisfying $2x^2-10x+10<0$.

Divide by 2: $x^2-5x+5<0$. Formula gives $x=\frac{5\pm\sqrt5}{2}\approx1.38$ or $3.62$.

Sketch: U-shape between these roots, solving "$<0$" (below axis) → $1.38<x<3.62$.

The only integers in this range are 2 and 3.

If the discriminant $b^2-4ac=0$, there's one repeated root where the vertex just touches the axis; if it's negative, the curve never crosses the axis at all. Either way, sketch first — the answer is often "all real numbers" or "no solution," decided entirely by whether the (U-shaped) graph ever dips below zero.

Example: $x^2-6x+13>0$ — discriminant $=36-52=-16<0$, so the graph never crosses the axis and stays entirely above it (its minimum is at $(3,4)$). Solution: all real numbers.

Solve the algebra first, then intersect with any physical constraints (e.g. side lengths must be positive).

Worked example: A rectangle has sides $(2x-4)$ and $(3x+2)$ cm, area $A<3x+27$. Find the range of $x$.

Algebra: $(2x-4)(3x+2)<3x+27 \implies 6x^2-11x-35<0$. Formula gives roots $x=3.5$ or $x=-\frac53$, so $-\frac53<x<3.5$.

Context: side $(2x-4)>0 \implies x>2$.

Intersecting both conditions: $2<x<3.5$.

Solve $x(x+2)\ge15$ and represent the solution on a number line.

Expand and set to standard form: $x^2+2x-15\ge0$. Factorise: $(x+5)(x-3)\ge0$, so critical values $x=-5,\ 3$.

Sketch: U-shape crossing at $-5$ and $3$. We want "$\ge0$" (on or above axis) → the two outer arms.

On a number line: closed circles at $-5$ and $3$ (non-strict), with arrows extending left from $-5$ and right from $3$.