Let the number be x.
The first part of the question can be written as
![x^2-10](https://img.qammunity.org/2023/formulas/mathematics/college/24rbx3wt9py0lapj8hhm6pmxf4whv25stv.png)
Adding the other part of the question, we have
![x^2-10=3x](https://img.qammunity.org/2023/formulas/mathematics/college/aypvh0an4esdvwsdjrm55w28kbsi1ttylr.png)
Rewriting the equation, we have
![x^2-3x-10=0](https://img.qammunity.org/2023/formulas/mathematics/college/632fmvcjsnz86lul9cno3ie8ialgqfmhp3.png)
Next, we solve the quadratic equation.
By factorization, we replace -3x with -5x and 2x. Hence
![\begin{gathered} x^2-5x+2x-10=0 \\ x(x-5)+2(x-5)=0 \\ (x-5)(x+2)=0 \\ \therefore \\ x=5\text{ or -2} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/2dynlvxct4v1323u6uvtm0tgvnkf6so2fq.png)
Therefore, the negative solution is -2.