You have to see if the polynomial is factorable
[tex]10x^2+5x-6=0[/tex]
In this case, it isn't, so now see the grade of the polynomial, we can see that is grade 2, so we can use the quadratic formula to find the roots:
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
the polynomial is in the form:
[tex]ax^2+bx+c[/tex]
So, replace
[tex]x=\frac{-5\pm\sqrt{5^2-(4*10*-6)}}{2*10}=\frac{-5\pm\sqrt{25+240}}{20}=\frac{-5\pm\sqrt{265}}{20}[/tex]
So we have two imaginary roots
[tex]x=\frac{-5+\sqrt{265}}{20},\:x=\frac{-5-\sqrt{265}}{20}[/tex]
