[tex]f(x) = 2x^2-10x-3 \\ \Delta = 100-4\cdot 2\cdot (-3) = 100+24 = 124 \\ \sqrt{\Delta} = \sqrt{124} = \sqrt{4\cdot31} = 2\sqrt{31}[/tex]
So the solutions of this equation are [tex]x_1 = \frac{10-2\sqrt{31}}{2\cdot2} = \frac{5-\sqrt{31}}{2}[/tex] and [tex]x_2 = \frac{5+\sqrt{31}}{2}[/tex].
