Answer:
[tex]x_1=\frac{1+\sqrt{33} }{2},y_1=\frac{5+\sqrt{33} }{2}\\ \\ x_2=\frac{1-\sqrt{33} }{2},y_2=\frac{5-\sqrt{33} }{2}[/tex]
Explanation:
You must find the solution of the system using the substitution method.
System:
[tex]y=-x^2+2x+10\\\\ y=x+2[/tex]
Substitute x+2 for y:
[tex]x+2=y=-x^2+2x+10[/tex]
Pass all the terms to one side:
[tex]x^2-2x+x+2-10\\ \\ x^2-x-8=0[/tex]
Use the quadratic formula to solve:
[tex]x=\frac{-b+/-\sqrt{b^2-4ac} }{2a}\\ \\ x=\frac{-(-1)+/-\sqrt{(-1)^2-4(1)(-8)} }{2(1)}\\ \\ x=\frac{1+/-\sqrt{1+32} }{2}[/tex]
[tex]x_1=\frac{1+\sqrt{33} }{2}\\ \\x_2=\frac{1-\sqrt{33} }{2}[/tex]
Use y = x + 2 to find the value of y:
[tex]y_1=2+x_1=2+\frac{1+\sqrt{33} }{2}\\ \\y_1=\frac{5+\sqrt{33} }{2} \\\\y_2=2+x_2=2+\frac{1-\sqrt{33} }{2}\\ \\ y_2=\frac{5-\sqrt{33} }{2}[/tex]
