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m^{2}-m-\frac{3}{4}=0
To solve the inequality, factor the left hand side. Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.
m=\frac{-\left(-1\right)±\sqrt{\left(-1\right)^{2}-4\times 1\left(-\frac{3}{4}\right)}}{2}
All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, -1 for b, and -\frac{3}{4} for c in the quadratic formula.
m=\frac{1±2}{2}
Do the calculations.
m=\frac{3}{2} m=-\frac{1}{2}
Solve the equation m=\frac{1±2}{2} when ± is plus and when ± is minus.
\left(m-\frac{3}{2}\right)\left(m+\frac{1}{2}\right)\geq 0
Rewrite the inequality by using the obtained solutions.
m-\frac{3}{2}\leq 0 m+\frac{1}{2}\leq 0
For the product to be ≥0, m-\frac{3}{2} and m+\frac{1}{2} have to be both ≤0 or both ≥0. Consider the case when m-\frac{3}{2} and m+\frac{1}{2} are both ≤0.
m\leq -\frac{1}{2}
The solution satisfying both inequalities is m\leq -\frac{1}{2}.
m+\frac{1}{2}\geq 0 m-\frac{3}{2}\geq 0
Consider the case when m-\frac{3}{2} and m+\frac{1}{2} are both ≥0.
m\geq \frac{3}{2}
The solution satisfying both inequalities is m\geq \frac{3}{2}.
m\leq -\frac{1}{2}\text{; }m\geq \frac{3}{2}
The final solution is the union of the obtained solutions.