Solve [2(m-7)]^2-4(m+3)(m-3)>0

$$[ 2 ( m - 7 ) ] ^ { 2 } - 4 ( m + 3 ) ( m - 3 ) > 0$$

Solve for m

$$m<\frac{29}{7}$$

Solution Steps

Use the distributive property to multiply 2 by m-7.

$$\left(2m-14\right)^{2}-4\left(m+3\right)\left(m-3\right)>0$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2m-14\right)^{2}.

$$4m^{2}-56m+196-4\left(m+3\right)\left(m-3\right)>0$$

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

$$4m^{2}-56m+196+\left(-4m-12\right)\left(m-3\right)>0$$

Use the distributive property to multiply -4m-12 by m-3 and combine like terms.

$$4m^{2}-56m+196-4m^{2}+36>0$$

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

$$-56m+196+36>0$$

Add 196 and 36 to get 232.

$$-56m+232>0$$

Subtract 232 from both sides. Anything subtracted from zero gives its negation.

$$-56m>-232$$

Divide both sides by -56. Since -56 is negative, the inequality direction is changed.

$$m<\frac{-232}{-56}$$

Reduce the fraction \frac{-232}{-56} to lowest terms by extracting and canceling out -8.

$$m<\frac{29}{7}$$