[tex]\begin{gathered} 3x^2-7x+10 \\ a=3 \\ b=-7 \\ c=10 \\ \text{Use the discriminant portion of the quadratic equation which is} \\ b^2-4ac \\ =(-7)^2-4(3)(10) \\ =49-120 \\ =-71 \\ \text{ Since the discriminant is negative, there are no real roots/solutions for the equation} \\ 3x^2-7x+10 \\ \text{ By standard quadratic formula, we have 2 non-real solutions for the equation which is} \\ x=\frac{7\pm\sqrt[]{-71}}{6} \end{gathered}[/tex]
