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

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

Solve for E

$$E=\left(2x-3\right)\left(4x-1\right)$$

Solution Steps

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

$$E=9x^{2}-12x+4-\left(x+1\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$E=9x^{2}-12x+4-\left(x^{2}+2x+1\right)$$

To find the opposite of x^{2}+2x+1, find the opposite of each term.

$$E=9x^{2}-12x+4-x^{2}-2x-1$$

Combine 9x^{2} and -x^{2} to get 8x^{2}.

$$E=8x^{2}-12x+4-2x-1$$

Combine -12x and -2x to get -14x.

$$E=8x^{2}-14x+4-1$$

Subtract 1 from 4 to get 3.

$$E=8x^{2}-14x+3$$

Solve for x (complex solution)

$$x=\frac{-\sqrt{8E+25}+7}{8}$$

$$x=\frac{\sqrt{8E+25}+7}{8}$$

Solve for x

$$x=\frac{-\sqrt{8E+25}+7}{8}$$

$$x=\frac{\sqrt{8E+25}+7}{8}\text{, }E\geq -\frac{25}{8}$$