Solve (2m+6)^2-4m^2+16m-12geq0

$$( 2 m + 6 ) ^ { 2 } - 4 m ^ { 2 } + 16 m - 12 \geq 0$$

Solve for m

$$m\geq -\frac{3}{5}$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2m+6\right)^{2}.

$$4m^{2}+24m+36-4m^{2}+16m-12\geq 0$$

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

$$24m+36+16m-12\geq 0$$

Combine 24m and 16m to get 40m.

$$40m+36-12\geq 0$$

Subtract 12 from 36 to get 24.

$$40m+24\geq 0$$

Subtract 24 from both sides. Anything subtracted from zero gives its negation.

$$40m\geq -24$$

Divide both sides by 40. Since 40 is positive, the inequality direction remains the same.

$$m\geq \frac{-24}{40}$$

Reduce the fraction \frac{-24}{40} to lowest terms by extracting and canceling out 8.

$$m\geq -\frac{3}{5}$$