Roots of $x^2-7x+12=0$ are

The given equation is $x^2-7x+12=0$

On comparing with standard form : $a{x}^2+bx+c=0$, we get a = 1, b = -7, c = 12

Now, $b^2-4ac={\left(-7\right)}^2-4\times 1\times 12=49-48=1$

we know that

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\therefore x=\frac{-\left(-7\right)\pm \sqrt{1}}{2\times 1}=\frac{7\pm 1}{2}$

Thus, $x=\frac{7-1}{2}=3 and x=\frac{7+1}{2}=4$

Hence, option D is correct