The other solution is correct. Here is an alternate
approach:

Since x² + x - 12 can be factored as (x+4)(x-3)
which has roots (zeros) x=-4 and x=3. Both must be 
roots (zeros) of x³ + ax² - 10x + b

So we substitute x=-4 in

   x³ +    ax² -    10x + b
(-4)³ + a(-4)² - 10(-4) + b
 -64  +    16a +     40 + b
              
and set that equal to 0

 -64  +  16a + 40 + b = 0
        -24 + 16a + b = 0
              16a + b = 24

We also substitute x=3 in

   x³ +   ax² -   10x + b
 (3)³ + a(3)² - 10(3) + b
   27  +   9a -    30 + b
              
and set that equal to 0

27  +  9a - 30 + b = 0
       -3 + 9a + b = 0
            9a + b = 3

Then we solve the system of equations;

16a + b = 24
 9a + b =  3

Solve that by substitution or elimination
and get

a=3 and b = -24    

Edwin