Solve (2-3x)(2+3x)+x(9x-2)=10

$$(2-3x)(2+3x)+x(9x-2)=10$$

Solve for x

$$x=-3$$

Steps for Solving Linear Equation

Consider \left(2-3x\right)\left(2+3x\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 2.

$$4-\left(3x\right)^{2}+x\left(9x-2\right)=10$$

Expand \left(3x\right)^{2}.

$$4-3^{2}x^{2}+x\left(9x-2\right)=10$$

Calculate 3 to the power of 2 and get 9.

$$4-9x^{2}+x\left(9x-2\right)=10$$

Use the distributive property to multiply x by 9x-2.

$$4-9x^{2}+9x^{2}-2x=10$$

Combine -9x^{2} and 9x^{2} to get 0.

$$4-2x=10$$

Subtract 4 from both sides.

$$-2x=10-4$$

Subtract 4 from 10 to get 6.

$$-2x=6$$

Divide both sides by -2.

$$x=\frac{6}{-2}$$

Divide 6 by -2 to get -3.

$$x=-3$$