jmc

We have $x\ge \lfloor x\rfloor >x-1,$ so $$x^2+x\ge 75>x^2+x-1.$$That is, $$75\le x^2+x<76.$$The function $f(x)=x^2+x$ is strictly decreasing for $x\le -1/2$; since $f(-10)=90$ and $f(-9)=72,$ it follows that any solution with $x\le -1/2$ must lie in the interval $(-10,-9).$ Similarly, since $f(8)=72$ and $f(9)=90,$ any solution with $x\ge -1/2$ must lie in the interval $(8,9).$

Therefore, $\lfloor x\rfloor$ can only be $-10$ or $8.$ If $\lfloor x\rfloor =-10,$ then $x^2=75-(-10)=85,$ so $x=-\sqrt{85},$ which indeed satisfies $\lfloor x\rfloor =-10.$ If $\lfloor x\rfloor =8,$ then $x^2=75-8=67,$ so $x=\sqrt{67},$ which indeed satisfies $\lfloor x\rfloor =67.$

Therefore, the two solutions to the equation are $x=\boxed{\sqrt{67},-\sqrt{85}}.$