[tex]f(x)=x^2-2x-8\\\\g(x)=x-1\\\\f(x)=g(x)\to x^2-2x-8=x-1\ \ \ |-x+1\\\\x^2-3x-7=0[/tex]
Use the quadratic formula:
[tex]ax^2+bx+c=0\\\\x_1=\dfrac{-b-\sqrt{b^2-4ac}}{2a};\x_2=\dfrac{-b+\sqrt{b^2-4ac}}{2a}[/tex]
[tex]x^2-3x-7=0\\\\a=1;\ b=-3;\ c=-7[/tex]
substitute
[tex]b^2-4ac=(-3)^2-4\cdot1\cdot(-7)=9+28=37\\\\\sqrt{b^2-4ac}=\sqrt{37}\approx6.083[/tex]
[tex]x_1=\dfrac{-(-3)-\sqrt{37}}{2\cdot1}=\dfrac{3-\sqrt{37}}{2}\approx\dfrac{3-6.083}{2}=-1.54\\\\x_2=\dfrac{-(-3)+\sqrt{37}}{2\cdot1}=\dfrac{3+\sqrt{37}}{2}\approx4.54[/tex]
there is no that answer to choose from
I also have such answers in the graphical method (look at the picture).
