Answer:
The values of x are
[tex]x1=\frac{5(+)\sqrt{97}}{12}[/tex]
[tex]x2=\frac{5(-)\sqrt{97}}{12}[/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]6x^{2} -5x-3=0[/tex]
so
[tex]a=6\\b=-5\\c=-3[/tex]
substitute in the formula
[tex]x=\frac{-(-5)(+/-)\sqrt{(-5)^{2}-4(6)(-3)}}{2(6)}[/tex]
[tex]x=\frac{5(+/-)\sqrt{97}}{12}[/tex]
[tex]x1=\frac{5(+)\sqrt{97}}{12}[/tex]
[tex]x2=\frac{5(-)\sqrt{97}}{12}[/tex]
