[tex]f(x) = log_4(x+1) + log_4(4x-3) = 0\\log_4\left((x+1)(4x-3)\right) = 0\\log_4\left(4x^2 +x -3 \right) = 0\\4x^2 +x -3 = 4^0 = 1\\4x^2 +x -4 = 0\\x = \dfrac{-1 \pm \sqrt{65}}{8}[/tex]
