Solve -3left(x-2right)^2+left(2x-1right)left(2x+1right)=left(4-xright)^2+4x+3

$$-3 { \left(x-2 \right) }^{ 2 } + \left( 2x-1 \right) \left( 2x+1 \right) = { \left(4-x \right) }^{ 2 } +4x+3$$

Solve for x

$$x=2$$

Steps for Solving Linear Equation

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

$$-3\left(x^{2}-4x+4\right)+\left(2x-1\right)\left(2x+1\right)=\left(4-x\right)^{2}+4x+3$$

Use the distributive property to multiply -3 by x^{2}-4x+4.

$$-3x^{2}+12x-12+\left(2x-1\right)\left(2x+1\right)=\left(4-x\right)^{2}+4x+3$$

Consider \left(2x-1\right)\left(2x+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

$$-3x^{2}+12x-12+\left(2x\right)^{2}-1=\left(4-x\right)^{2}+4x+3$$

Expand \left(2x\right)^{2}.

$$-3x^{2}+12x-12+2^{2}x^{2}-1=\left(4-x\right)^{2}+4x+3$$

Calculate 2 to the power of 2 and get 4.

$$-3x^{2}+12x-12+4x^{2}-1=\left(4-x\right)^{2}+4x+3$$

Combine -3x^{2} and 4x^{2} to get x^{2}.

$$x^{2}+12x-12-1=\left(4-x\right)^{2}+4x+3$$

Subtract 1 from -12 to get -13.

$$x^{2}+12x-13=\left(4-x\right)^{2}+4x+3$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4-x\right)^{2}.

$$x^{2}+12x-13=16-8x+x^{2}+4x+3$$

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

$$x^{2}+12x-13=16-4x+x^{2}+3$$

Add 16 and 3 to get 19.

$$x^{2}+12x-13=19-4x+x^{2}$$

Add 4x to both sides.

$$x^{2}+12x-13+4x=19+x^{2}$$

Combine 12x and 4x to get 16x.

$$x^{2}+16x-13=19+x^{2}$$

Subtract x^{2} from both sides.

$$x^{2}+16x-13-x^{2}=19$$

Combine x^{2} and -x^{2} to get 0.

$$16x-13=19$$

Add 13 to both sides.

$$16x=19+13$$

Add 19 and 13 to get 32.

$$16x=32$$

Divide both sides by 16.

$$x=\frac{32}{16}$$

Divide 32 by 16 to get 2.

$$x=2$$