Solve F=(x+2)^2-(3-2x)(3+2x)

$$F = ( x + 2 ) ^ { 2 } - ( 3 - 2 x ) ( 3 + 2 x )$$

Solve for F

$$F=5x^{2}+4x-5$$

Solution Steps

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

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

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

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

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

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

Calculate 2 to the power of 2 and get 4.

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

To find the opposite of 9-4x^{2}, find the opposite of each term.

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

Subtract 9 from 4 to get -5.

$$F=x^{2}+4x-5+4x^{2}$$

Combine x^{2} and 4x^{2} to get 5x^{2}.

$$F=5x^{2}+4x-5$$

Solve for x (complex solution)

$$x=\frac{-\sqrt{5F+29}-2}{5}$$

$$x=\frac{\sqrt{5F+29}-2}{5}$$

Solve for x

$$x=\frac{-\sqrt{5F+29}-2}{5}$$

$$x=\frac{\sqrt{5F+29}-2}{5}\text{, }F\geq -\frac{29}{5}$$