Question

Find the real zeros 4x²-16x + 2.5x​

Answer 1

I am going to assume 2.5x is really 25x (typo).

First formulate the equality

$4x^2-16x+25x=0 \\4x^2+9x=0 \\x(4x+9)=0$

Now two results emerge.

$x_1=0, x_2=-\dfrac{9}{4}$.

Hope this helps.