Answer:
[tex]x1=\frac{5+\sqrt{37}} {6}[/tex]
[tex]x2=\frac{5-\sqrt{37}} {6}[/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(+/-)\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]3x^{2} -5x-1=0[/tex]
so
[tex]a=3\\b=-5\\c=-1[/tex]
substitute in the formula
[tex]x=\frac{5(+/-)\sqrt{(-5)^{2}-4(3)(-1)}} {2(3)}[/tex]
[tex]x=\frac{5(+/-)\sqrt{37}} {6}[/tex]
[tex]x1=\frac{5+\sqrt{37}} {6}[/tex]
[tex]x2=\frac{5-\sqrt{37}} {6}[/tex]
