Solve (7-x)^2+41=x^2

$$( 7 - x ) ^ { 2 } + 41 = x ^ { 2 }$$

Solve for x

$$x = \frac{45}{7} = 6\frac{3}{7} \approx 6.428571429$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(7-x\right)^{2}.

$$49-14x+x^{2}+41=x^{2}$$

Add 49 and 41 to get 90.

$$90-14x+x^{2}=x^{2}$$

Subtract x^{2} from both sides.

$$90-14x+x^{2}-x^{2}=0$$

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

$$90-14x=0$$

Subtract 90 from both sides. Anything subtracted from zero gives its negation.

$$-14x=-90$$

Divide both sides by -14.

$$x=\frac{-90}{-14}$$

Reduce the fraction \frac{-90}{-14} to lowest terms by extracting and canceling out -2.

$$x=\frac{45}{7}$$