Answer:
[tex](\frac{19-\sqrt{365}} {2},\frac{19+\sqrt{365}} {2})[/tex]
Step-by-step explanation:
we have
[tex]x^2=19x+1[/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\pm\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]-x^{2}-19x-1=0[/tex]
so
[tex]a=1\\b=-19\\c=-1[/tex]
substitute in the formula
[tex]x=\frac{-(-19)\pm\sqrt{-19^{2}-4(1)(-1)}} {2(1)}[/tex]
[tex]x=\frac{19\pm\sqrt{365}} {2}[/tex]
[tex]x=\frac{19+\sqrt{365}} {2}[/tex]
[tex]x=\frac{19-\sqrt{365}} {2}[/tex]
[tex](\frac{19-\sqrt{365}} {2},\frac{19+\sqrt{365}} {2})[/tex]
therefore
StartFraction 19 minus StartRoot 365 EndRoot Over 2 EndFraction comma StartFraction 19 + StartRoot 365 EndRoot Over 2 EndFraction
