In solving the equation $x^2+ax+b=0$ wrong value of ‘$a$’ was taken and the roots were found to be 6 and 2. If again a wrong value of b was taken and the roots were found to be -2 and 9 then the correct roots of the original equation are

Given that the roots of the equation.$x^2+ax+b=0$ are $6,2$, when the value of $a$ is wrong.

i.e.,  the product of the roots is $b=12$ 

Given that the roots of the equation $x^2+ax+b=0$ are $-2,9$, when the value of $b$ is wrong

i.e., the sum of the roots is $a=-\left(-2+9\right)=-7$

Therefore, the original equation is $x^2-7x+12=0$

On factorizing, the roots of above equation are  $\boxed{\mathbf{3}\mathbf{,}\mathbf{4}}$