Solve the inequality (-m)^2 - 4(-m + 3) > 0.

Step-by-step solution

1. 1

   Simplify the squared term. An even power destroys the sign: $(-m)^2=(-1)^2m^2=m^2$. Writing $-m^2$ instead is a frequent slip that changes the whole problem, because it would flip the parabola.

2. 2

   Distribute the -4 across the parentheses. Both inner signs change:

   $-4\cdot(-m)=+4m,\qquad -4\cdot 3=-12$

   so the inequality becomes

   $m^2+4m-12>0$

   (This shape often arises as the discriminant condition $b^2-4ac>0$ for a quadratic to have two distinct real roots.)

3. 3

   Factor the trinomial. Two numbers multiplying to $-12$ and adding to $+4$ are $+6$ and $-2$:

   $m^2+4m-12=(m+6)(m-2)$

   so the inequality is $(m+6)(m-2)>0$.

4. 4

   Read the sign of the product on each interval. The critical points are $m=-6$ and $m=2$. For $m<-6$ both factors are negative, so the product is positive. For $-6<m<2$ the first is positive and the second negative, so the product is negative. For $m>2$ both are positive, so the product is positive again.

5. 5

   Select the intervals where the product is positive, and exclude the roots. Because the inequality is strict, $m=-6$ and $m=2$ — where the product is exactly $0$ — are not included:

   $m<-6\qquad\text{or}\qquad m>2$

6. 6

   Verify the factorisation and the solution set. The factored form matches the expanded one at $401$ sample values of $m$ ✓, and scanning $2401$ points from $-12$ to $12$ confirms the sign pattern agrees with $(-\infty,-6)\cup(2,\infty)$ at every point ✓.