Solve left(3x-5right)^2+left(7+2xright)^2=left(2x-3right)^2+9x^2+65

$${ \left(3x-5 \right) }^{ 2 } + { \left(7+2x \right) }^{ 2 } = { \left(2x-3 \right) }^{ 2 } +9 { x }^{ 2 } +65$$

Solve for x

$$x=0$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-5\right)^{2}.

$$9x^{2}-30x+25+\left(7+2x\right)^{2}=\left(2x-3\right)^{2}+9x^{2}+65$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(7+2x\right)^{2}.

$$9x^{2}-30x+25+49+28x+4x^{2}=\left(2x-3\right)^{2}+9x^{2}+65$$

Add 25 and 49 to get 74.

$$9x^{2}-30x+74+28x+4x^{2}=\left(2x-3\right)^{2}+9x^{2}+65$$

Combine -30x and 28x to get -2x.

$$9x^{2}-2x+74+4x^{2}=\left(2x-3\right)^{2}+9x^{2}+65$$

Combine 9x^{2} and 4x^{2} to get 13x^{2}.

$$13x^{2}-2x+74=\left(2x-3\right)^{2}+9x^{2}+65$$

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

$$13x^{2}-2x+74=4x^{2}-12x+9+9x^{2}+65$$

Combine 4x^{2} and 9x^{2} to get 13x^{2}.

$$13x^{2}-2x+74=13x^{2}-12x+9+65$$

Add 9 and 65 to get 74.

$$13x^{2}-2x+74=13x^{2}-12x+74$$

Subtract 13x^{2} from both sides.

$$13x^{2}-2x+74-13x^{2}=-12x+74$$

Combine 13x^{2} and -13x^{2} to get 0.

$$-2x+74=-12x+74$$

Add 12x to both sides.

$$-2x+74+12x=74$$

Combine -2x and 12x to get 10x.

$$10x+74=74$$

Subtract 74 from both sides.

$$10x=74-74$$

Subtract 74 from 74 to get 0.

$$10x=0$$

Product of two numbers is equal to 0 if at least one of them is 0. Since 10 is not equal to 0, x must be equal to 0.

$$x=0$$