Question

What are the zeros of the quadratic function f(x) = 6x2 – 24x + 1?

Answer 1

Answer:

$x=\frac{12+\sqrt{138}}{6}\:\:or\:\:x=\frac{12-\sqrt{138}}{6}$

Step-by-step explanation:

$f(x)=6x^2-24x+1\\\\0=6x^2-24x+1\\\\x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\x=\frac{-(-24)\pm\sqrt{(-24)^2-4(6)(1)}}{2(6)}\\ \\x=\frac{24\pm\sqrt{576-24}}{12}\\\\x=\frac{24\pm2\sqrt{138}}{12}\\\\x=\frac{12\pm\sqrt{138}}{6}$