Let xbe a positive integer number. Then x, x+1 and x+2 are three positive consecutive integers (the first one is x, the second is x+1 and the third is x+2).
The product of the second integer and the third integer is (x+1)·(x+2) and is equal to 72. So you have the equation
[tex](x+1)\cdot (x+2)=72.[/tex]
Solve it:
[tex]x^2+2x+x+2=72,\\ \\x^2+3x+2-72=0,\\ \\x^2+3x-70=0,\\ \\D=3^2-4\cdot (-70)=9+280=289,\\ \\\sqrt{D}17,\\ \\x_1=\dfrac{-3-17}{2}=-10,\ x_2=\dfrac{-3+17}{2}=7.[/tex]
Solution [tex]x_1=-10[/tex] is extra because [tex]x_1[/tex] is negative.
Answer: three positive consecutive integers are 7, 8 and 9.
