Solve left(x+4right)^2-(x-5)(x+5)=0

$${ \left(x+4 \right) }^{ 2 } -(x-5)(x+5)=0$$

Solve for x

$$x = -\frac{41}{8} = -5\frac{1}{8} = -5.125$$

Steps for Solving Linear Equation

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+4\right)^{2}.

$$x^{2}+8x+16-\left(x-5\right)\left(x+5\right)=0$$

Consider \left(x-5\right)\left(x+5\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 5.

$$x^{2}+8x+16-\left(x^{2}-25\right)=0$$

To find the opposite of x^{2}-25, find the opposite of each term.

$$x^{2}+8x+16-x^{2}+25=0$$

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

$$8x+16+25=0$$

Add 16 and 25 to get 41.

$$8x+41=0$$

Subtract 41 from both sides. Anything subtracted from zero gives its negation.

$$8x=-41$$

Divide both sides by 8.

$$x=\frac{-41}{8}$$

Fraction \frac{-41}{8} can be rewritten as -\frac{41}{8} by extracting the negative sign.

$$x=-\frac{41}{8}$$