Solve 16-x^2+(4-x)^2=28

$$16 - x ^ { 2 } + ( 4 - x ) ^ { 2 } = 28$$

Solve for x

$$x=\frac{1}{2}=0.5$$

Steps for Solving Linear Equation

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

$$16-x^{2}+16-8x+x^{2}=28$$

Add 16 and 16 to get 32.

$$32-x^{2}-8x+x^{2}=28$$

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

$$32-8x=28$$

Subtract 32 from both sides.

$$-8x=28-32$$

Subtract 32 from 28 to get -4.

$$-8x=-4$$

Divide both sides by -8.

$$x=\frac{-4}{-8}$$

Reduce the fraction \frac{-4}{-8} to lowest terms by extracting and canceling out -4.

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