we have x+y=5 xy=-45 solve x+y=5 subtract x from both sides y=-x+5 subsiute -x+5 for y in second equaiton x(-x+5)=-45 -x^2+5x=-45 add 2x^2-5x to both sides 0=x^2-5x-45 use quadratic equation which is if you have 0=ax^2+bx+c then x=[tex] \frac{-b+ \sqrt{b^{2}-4ac} }{2a} [/tex] or [tex] \frac{-b- \sqrt{b^{2}-4ac} }{2a} [/tex] sbsitute 1x^2-5x-45 a=1 b=-5 c=-45 subsitute [tex] \frac{-(-5)+ \sqrt{-5^{2}-4(1)(-45)} }{2(1)} [/tex] [tex] \frac{5+ \sqrt{25-(-180)} }{2} [/tex] [tex] \frac{5+ \sqrt{25+180} }{2} [/tex] [tex] \frac{5+ \sqrt{205} }{2} [/tex] [tex] \sqrt{205} [tex} is not eaqual to a whole number so we will leave it like that
x= [tex] \frac{5+ \sqrt{205} }{2} [/tex] or[tex] \frac{5- \sqrt{} }{2} [/tex] or aprox x=-4.65891 or 9.65891, so the 2 numbers would be those answers for x
