We have the next equation:

Substituting with:

we get:

Applying the quadratic formula to the last equation:
![\begin{gathered} y_(1,2)=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_(1,2)=\frac{-(-1)\pm\sqrt[]{(-1)^2-4\cdot1\cdot(-12)}}{2\cdot1} \\ y_(1,2)=(1\pm7)/(2) \\ y_1=(1+7)/(2)=4 \\ y_2=(1-7)/(2)=-3 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/nbae8ja1187894n50y0ajmny8192q9dh14.png)
Coming back to the original variable:
![\begin{gathered} y_{}=x^2 \\ 4=x^2 \\ \sqrt[]{4}=x \\ \text{This equation has two solutions:} \\ 2=x_1 \\ -2=x_2 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/czqomat8vwntlssbn4ozm8wtfz84we5to8.png)
And with the other solution:
![\begin{gathered} y=x^2 \\ -3=x^2 \\ \sqrt[]{-3}=x \\ \sqrt[]{3\cdot(-1)}=x \\ \sqrt[]{3}\cdot\sqrt[]{-1}=x \\ \sqrt[]{3}i=x \\ \text{This equation has two solutions:} \\ 0+\sqrt[]{3}i=x_3 \\ 0-\sqrt[]{3}i=x_4 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/jojfde4sipb5qecq23r2jbwx139vurseyn.png)
In conclusion, there are two unequal real roots (x1 and x2) and two unequal complex roots (x3 and x4).