Answer:
[tex]x=\frac{2}{11}\pm\frac{\sqrt{15}} {11}[/tex]
Step-by-step explanation:
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]11x^{2} -4x=1[/tex]
equate to zero
[tex]11x^{2} -4x-1=0[/tex]
so
[tex]a=11\\b=-4\\c=-1[/tex]
substitute in the formula
[tex]x=\frac{-(-4)\pm\sqrt{-4^{2}-4(11)(-1)}} {2(11)}[/tex]
[tex]x=\frac{4\pm\sqrt{60}} {22}[/tex]
[tex]x=\frac{4\pm2\sqrt{15}} {22}[/tex]
[tex]x=\frac{2\pm\sqrt{15}} {11}[/tex]
[tex]x=\frac{2}{11}\pm\frac{\sqrt{15}} {11}[/tex]
therefore
StartFraction 2 Over 11 EndFraction plus-or-minus StartFraction StartRoot 15 EndRoot Over 11 EndFraction
