Solve [2(a-1)]^2-4(a^2-a-2)=0

$$[ 2 ( a - 1 ) ] ^ { 2 } - 4 ( a ^ { 2 } - a - 2 ) = 0$$

Solve for a

$$a=3$$

Steps for Solving Linear Equation

Use the distributive property to multiply 2 by a-1.

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

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.

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

Divide -12 by -4 to get 3.

$$a=3$$