Help Tom!

The quadratic equation $x^2+3x+2=0$ factorizes to $(x+2)(x+1)=0$ which implies that $x=-1$ or $x=-2$ . It is obvious that $-1$ is closer to $0$ (plot it on the number line). Therefore, $\boxed{x=-1}$

This is a quadratic equation.

$x^2+3x+2=0$

We can use the quadratic equation to solve this question.

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

If we plug in the values this will be formed:

$x=\frac{-3\pm\sqrt{3^2-4(1)(2)}}{2}.$

This can be further simplified as:

$x=\frac{-3\pm1}{2}.$

We can then find both x values.

So first for $x_1$ . It becomes like the following:

$x_1=\frac{-3+1}{2}=-1.$

And for $x_2$ . It will become

$x_2=\frac{-3-1}{2}=-2.$

And then it says us to find the value of x which is the closest to 0. In that case, more $x_1$ more closer which is:

$-1.$

So the answer to this question is

= $-1.$