Answer:
The answer in the procedure
Step-by-step explanation:
we have
[tex]2=-x+x^{2} -4[/tex]
[tex]x^{2}-x-4-2=0[/tex]
[tex]x^{2}-x-6=0[/tex]
we know that
The formula to solve a quadratic equation of the form
[tex]ax^{2} +bx+c=0[/tex]
is equal to
[tex]x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]x^{2}-x-6=0[/tex]
so
[tex]a=1\\b=-1\\c=-6[/tex]
substitute in the formula
[tex]x=\frac{-(-1)(+/-)\sqrt{(-1)^{2}-4(1)(-6)}} {2(1)}[/tex]
[tex]x=\frac{1(+/-)\sqrt{25}} {2}[/tex]
[tex]x=\frac{1(+/-)5} {2}[/tex]
[tex]x=\frac{1(+)5} {2}=3[/tex]
[tex]x=\frac{1(-)5} {2}=-2[/tex]
