You can divide the whole equation by 2 to get
[tex] x^2 -\frac{1}{2}x-5=0 [/tex]
This is useful, in this case, because any quadratic equation with leading coefficient 1 can be read as
[tex] x^2-sx+p = 0 [/tex]
where s is the sum of the two solutions, and p is their product.
So, we know that the two solutions sum to 1/2, and give -5 when multiplied.
But we already know that one solution is 5/2, so if we call the other solution x we have
[tex] \frac{5}[2}+x = \frac{1}{2} \iff x = \frac{1}{2} - \frac{5}[2} = \frac{-4}{2} = -2 [/tex]
Just to check, let's see that they give -5 when multiplied:
[tex] \frac{5}[2} \cdot (-2) = -5 [/tex]
So it works! The other solution is -2.
