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Show that every solution of the constants coefficient equation $y''+a_1y'+a_2y=0$ tends to zero as $x→∞$ if, and only if, the real parts of the roots of the characteristic polynomial are negative.

Using quadratic equation I found: $$x = \frac{-a_1}{2} ±\frac{\sqrt{a_1^2-4a_2}}{2}$$

Please help me with this. Thank you.

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    If the roots are $r,s$ then the solutions are of the form $f(x)=ae^{rx}+be^{sx}$. It's not necessary to solve for $r,s$ in terms of $a_1,a_2$. Now, what happens as $x\to \infty$? – 2017-02-20

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If the roots $m_1$ and $m_2$ are real and negative then $lim_{x\rightarrow \infty}y(x)=A.e^{m_1.x}+B.e^{m_2.x}\rightarrow 0$.

If the roots are complex i.e. $m_1=a+ib$ and $m_2=a-ib$ with $a<0$ , then $lim_{x\rightarrow \infty}y(x)=e^{a.x}[Acos(bx)+Bsin(bx)]\rightarrow 0$.