Question

$x^2-7x-18=0,x=\sqrt{3},-2,4$

Answer 1

Given equation is:

$x^2-7x-18=0$
$x^2-9x+2x-18=0$
$(x-9)(x+2)=0$
$\therefore x=9,-2$
Hence, $\sqrt{3}$, $4$ are not the roots of the given equation.