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) .