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

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

Solve for m

$$m<-\frac{11}{20}$$

Solution Steps

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

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

Use the distributive property to multiply -4 by m+1.

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

Use the distributive property to multiply -4m-4 by m+3 and combine like terms.

$$4m^{2}-4m+1-4m^{2}-16m-12>0$$

Combine 4m^{2} and -4m^{2} to get 0.

$$-4m+1-16m-12>0$$

Combine -4m and -16m to get -20m.

$$-20m+1-12>0$$

Subtract 12 from 1 to get -11.

$$-20m-11>0$$

Add 11 to both sides. Anything plus zero gives itself.

$$-20m>11$$

Divide both sides by -20. Since -20 is negative, the inequality direction is changed.

$$m<-\frac{11}{20}$$