Question

Thanks for the help, question below :)

Answer 1

Hello, please consider the following.

$x^2-6x=-5\\\\\text{We add 5 to both sides of the equation}\\\\x^2-6x+\boxed{5}=-5+\boxed{5}=0\\\\\text{The answer to a) is then c = 5.}\\\\\text{We complete the square by noticing that }\\\\x^2-6x=x^2-2*3*x=(x-3)^2-3^2\\\\\text{So the main equation becomes.}\\\\x^2-6x+5=(x-3)^2-9+5=0\\\\\text{We simplify.}\\\\(x-3)^2-4=0$

$\text{We add 4 to both sides.}\\\\(x-3)^2=4=2^2\\\\\text{The answer to b) is then}\\\\\boxed{(x-3)^2=4}\\\\\text{We take the root.}\\\\x-3=\pm2\\\\ x =3+2=5 \ \ or \ \ x=3-2=1\\\\\text{So the solutions are.}\\\\\Large \boxed{\sf \bf \ \ x=5 \ \ or \ \ x=1 \ \ }$

Thank you.