[tex]f(x) = 4x^{2}+x-3\\\\f(x) =0\\\\4x^{2}+x-3 = 0\\\\a = 4, \ b = 1, \ c = -1\\\\\Delta = b^{2}-4ac = 1^{2}-4\cdot4\cdot(-3) =1+48 = 49\\\\\sqrt{\Delta} = \sqrt{49} = 7\\\\x_1 = \frac{-b-\sqrt{\Delta}}{2a} = \frac{-1-7}{2\cdot4} = \frac{-8}{8} = -1\\\\x_2 = \frac{-b+\sqrt{\Delta}}{2a} = \frac{-1+7}{8} = \frac{6}{8} = \frac{3}{4}\\\\x \in\{-1, \frac{3}{4}\}[/tex]
