a) x^2-4x+5=0

b) x^2+7x+12=0

c) 2x^2-7x+6=0

d) 3x^2-6x-2=0

correct answer is: b) x^2+7x+12=0

**Explanation**

Let’s substitute ‘-3’ into each equation:** (a)** (x^{2} - 4x + 5 = 0):

\rightarrow\left(-3\right)^2-4\times\left(-3\right)+5

\rightarrow9+12+5

\rightarrow26

\therefore26\neq0

** (b)** (x^{2} + 7x + 12 = 0):

\rightarrow\left(-3\right)^2+7\times\left(-3\right)+12

\rightarrow9-21+12

\rightarrow21-21

\rightarrow0

\therefore0=0

** (c)** (2x^{2} - 7x + 6 = 0):

\rightarrow2\times\left(-3\right)^2-7\times\left(-3\right)+6

\rightarrow2\times9+21+6

\rightarrow18+21+6

\rightarrow45

\therefore45\neq0

** (d)** (3x^{2} - 6x - 2 = 0):

\rightarrow3\times\left(-3\right)^2-6\times\left(-3\right)-2

\rightarrow3\times9+18-2

\rightarrow27+18-2

\rightarrow43

\therefore43\neq0

So, the equation (x^{2} + 7x + 12 = 0) has -3 as a root.

**Ans:**Correct answer is (x^{2} + 7x + 12 = 0) .