Solve (2m-1)^2-3(m+4)<0

$$( 2 m - 1 ) ^ { 2 } - 3 ( m + 4 ) < 0$$

Solve for m

$$m\in \left(-1,\frac{11}{4}\right)$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2m-1\right)^{2}.

$$4m^{2}-4m+1-3\left(m+4\right)<0$$

Use the distributive property to multiply -3 by m+4.

$$4m^{2}-4m+1-3m-12<0$$

Combine -4m and -3m to get -7m.

$$4m^{2}-7m+1-12<0$$

Subtract 12 from 1 to get -11.

$$4m^{2}-7m-11<0$$

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.

$$4m^{2}-7m-11=0$$

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 4 for a, -7 for b, and -11 for c in the quadratic formula.

$$m=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 4\left(-11\right)}}{2\times 4}$$

Do the calculations.

$$m=\frac{7±15}{8}$$

Solve the equation m=\frac{7±15}{8} when ± is plus and when ± is minus.

$$m=\frac{11}{4} m=-1$$

Rewrite the inequality by using the obtained solutions.

$$4\left(m-\frac{11}{4}\right)\left(m+1\right)<0$$

For the product to be negative, m-\frac{11}{4} and m+1 have to be of the opposite signs. Consider the case when m-\frac{11}{4} is positive and m+1 is negative.

$$m-\frac{11}{4}>0 m+1<0$$

This is false for any m.

$$m\in \emptyset$$

Consider the case when m+1 is positive and m-\frac{11}{4} is negative.

$$m+1>0 m-\frac{11}{4}<0$$

The solution satisfying both inequalities is m\in \left(-1,\frac{11}{4}\right).

$$m\in \left(-1,\frac{11}{4}\right)$$

The final solution is the union of the obtained solutions.

$$m\in \left(-1,\frac{11}{4}\right)$$