Question 12429
What you did so far, looks right so far.  Let's expand the square of the binomial:


y=1-2x
x²+(1-2x)²=13 

{{{x^2 + 1 - 4x + 4x^2 = 13}}}
{{{5x^2 - 4x - 12= 0}}}

By quadratic formula
{{{x = (4 +- sqrt(16-4*5*(-12)))/(2*5)}}}
{{{x = (4 +- sqrt(16+240))/10}}}
{{{x = (4 +- sqrt(256))/10}}}


This is the square root of a perfect square, which means that the original equation DOES factor, and the quadratic formula was not necessary in this case.  However, you are almost there now, so might as well finish it.

{{{x= (4 +- 16)/10}}} 


{{{x= (4+16)/10 = 20/10 = 2}}}
{{{x= (4-16)/10 = (-12)/10 = (-6)/5}}}


Then solve for y using y = 1-2x
If x= 2, y = 1-2*2= -3


If {{{x = -6/5}}}, then {{{y = 1 - 2*((-6)/5)}}}= {{{1+ 12/5}}} = {{{17/5}}}


Points of solution would be  (2,-3)  and ({{{  (-6)/5}}}, {{{17/5}}} ).


Check for errors in my work!!  Let me know if I made any.


R^2 at SCC