Solve 16x^2+(2sqrt{3}-3x)^2-4x(2sqrt{3}-3x)=(2sqrt{3}-x)^2

$$16 x ^ { 2 } + ( 2 \sqrt { 3 } - 3 x ) ^ { 2 } - 4 x ( 2 \sqrt { 3 } - 3 x ) = ( 2 \sqrt { 3 } - x ) ^ { 2 }$$

Solve for x

$$x=\frac{4\sqrt{3}}{9}\approx 0.769800359$$

$$x=0$$

Steps Using Factoring

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2\sqrt{3}-3x\right)^{2}.

$$16x^{2}+4\left(\sqrt{3}\right)^{2}-12\sqrt{3}x+9x^{2}-4x\left(2\sqrt{3}-3x\right)=\left(2\sqrt{3}-x\right)^{2}$$

The square of \sqrt{3} is 3.

$$16x^{2}+4\times 3-12\sqrt{3}x+9x^{2}-4x\left(2\sqrt{3}-3x\right)=\left(2\sqrt{3}-x\right)^{2}$$

Multiply 4 and 3 to get 12.

$$16x^{2}+12-12\sqrt{3}x+9x^{2}-4x\left(2\sqrt{3}-3x\right)=\left(2\sqrt{3}-x\right)^{2}$$

Combine 16x^{2} and 9x^{2} to get 25x^{2}.

$$25x^{2}+12-12\sqrt{3}x-4x\left(2\sqrt{3}-3x\right)=\left(2\sqrt{3}-x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2\sqrt{3}-x\right)^{2}.

$$25x^{2}+12-12\sqrt{3}x-4x\left(2\sqrt{3}-3x\right)=4\left(\sqrt{3}\right)^{2}-4\sqrt{3}x+x^{2}$$

The square of \sqrt{3} is 3.

$$25x^{2}+12-12\sqrt{3}x-4x\left(2\sqrt{3}-3x\right)=4\times 3-4\sqrt{3}x+x^{2}$$

Multiply 4 and 3 to get 12.

$$25x^{2}+12-12\sqrt{3}x-4x\left(2\sqrt{3}-3x\right)=12-4\sqrt{3}x+x^{2}$$

Subtract 12 from both sides.

$$25x^{2}+12-12\sqrt{3}x-4x\left(2\sqrt{3}-3x\right)-12=-4\sqrt{3}x+x^{2}$$

Add 4\sqrt{3}x to both sides.

$$25x^{2}+12-12\sqrt{3}x-4x\left(2\sqrt{3}-3x\right)-12+4\sqrt{3}x=x^{2}$$

Subtract x^{2} from both sides.

$$25x^{2}+12-12\sqrt{3}x-4x\left(2\sqrt{3}-3x\right)-12+4\sqrt{3}x-x^{2}=0$$

Use the distributive property to multiply -4x by 2\sqrt{3}-3x.

$$25x^{2}+12-12\sqrt{3}x-8\sqrt{3}x+12x^{2}-12+4\sqrt{3}x-x^{2}=0$$

Combine -12\sqrt{3}x and -8\sqrt{3}x to get -20\sqrt{3}x.

$$25x^{2}+12-20\sqrt{3}x+12x^{2}-12+4\sqrt{3}x-x^{2}=0$$

Combine 25x^{2} and 12x^{2} to get 37x^{2}.

$$37x^{2}+12-20\sqrt{3}x-12+4\sqrt{3}x-x^{2}=0$$

Subtract 12 from 12 to get 0.

$$37x^{2}-20\sqrt{3}x+4\sqrt{3}x-x^{2}=0$$

Combine -20\sqrt{3}x and 4\sqrt{3}x to get -16\sqrt{3}x.

$$37x^{2}-16\sqrt{3}x-x^{2}=0$$

Combine 37x^{2} and -x^{2} to get 36x^{2}.

$$36x^{2}-16\sqrt{3}x=0$$

Factor out x.

$$x\left(36x-16\sqrt{3}\right)=0$$

To find equation solutions, solve x=0 and 36x-16\sqrt{3}=0.

$$x=0 x=\frac{4\sqrt{3}}{9}$$