The question says that the product of two consecutives even integers is 120, wich give us the expression:
[tex]x(x+2)=120[/tex]Developing the calculations, we end with the followed second degree equation:
[tex]x^2\text{ + 2x = 120 }\Rightarrow\text{ x}^2\text{ + 2x - 120=0}[/tex]We can use Bhaskara's formula to find the roots, as follow:
[tex]x=\frac{-2\pm\sqrt[]{2^2-4(1)(-120)}}{2\times1}\text{ = }\frac{-2\pm\sqrt[]{4+480}}{2}[/tex]Continuing the calculos, we find:
[tex]x=\frac{-2\pm\sqrt[]{484}}{2}=\text{ }\frac{-2\pm22}{2}[/tex]So, our x can be:
[tex]\frac{-2-22}{2}\text{ = -12 or }\frac{-2+22}{2}=\text{ 10}[/tex]We want the positive solution, so our integers are 10 and 12.
