Solve left(-2a+2right)^2-4left(a^2-a-2right)>0

$${ \left(-2a+2 \right) }^{ 2 } -4 \left( { a }^{ 2 } -a-2 \right) > 0$$

Solve for a

$$a<3$$

Solution Steps

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

$$4a^{2}-8a+4-4\left(a^{2}-a-2\right)>0$$

Use the distributive property to multiply -4 by a^{2}-a-2.

$$4a^{2}-8a+4-4a^{2}+4a+8>0$$

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

$$-8a+4+4a+8>0$$

Combine -8a and 4a to get -4a.

$$-4a+4+8>0$$

Add 4 and 8 to get 12.

$$-4a+12>0$$

Subtract 12 from both sides. Anything subtracted from zero gives its negation.

$$-4a>-12$$

Divide both sides by -4. Since -4 is negative, the inequality direction is changed.

$$a<\frac{-12}{-4}$$

Divide -12 by -4 to get 3.

$$a<3$$