x 2 + 5 x + 6 = 0

 D = 

$b^2–4ac$

D = $5^2-4\times 1\times 6$ = $25–24$ = $1$

Since D>0, the equation will have two real and distinct roots. The roots are:

$x$ = $\frac{-b+\surd D}{2a}$ or $\frac{-b–\surd D}{2a}$

$x$ = $\frac{-5+\surd 1}{2\times 1}$ or $\frac{-5–\surd 1}{2\times 1}$

$x$ = $-2$ or $-3$