Solve (x-4)^2-(x+1)^2=0

$$( x - 4 ) ^ { 2 } - ( x + 1 ) ^ { 2 } = 0$$

Solve for x

$$x = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

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}.

$$x^{2}-8x+16-\left(x+1\right)^{2}=0$$

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

$$x^{2}-8x+16-\left(x^{2}+2x+1\right)=0$$

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

$$x^{2}-8x+16-x^{2}-2x-1=0$$

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

$$-8x+16-2x-1=0$$

Combine -8x and -2x to get -10x.

$$-10x+16-1=0$$

Subtract 1 from 16 to get 15.

$$-10x+15=0$$

Subtract 15 from both sides. Anything subtracted from zero gives its negation.

$$-10x=-15$$

Divide both sides by -10.

$$x=\frac{-15}{-10}$$

Reduce the fraction \frac{-15}{-10} to lowest terms by extracting and canceling out -5.

$$x=\frac{3}{2}$$