12.50 Find the equations of the tangents to the curves $y=2 x^{2}-5$ and $y=x^{2}-3 x+5$, drawn through the points of intersection of these curves.
12.50 To find the points of intersection of the given curves, we solve the system of equations

$$
\left\{\begin{array}{l}
y=2 x^{2}-5 \\
y=x^{2}-3 x+5
\end{array}\right.
$$

from which we obtain $x_{1}=-5, x_{2}=2$ and $y_{1}=45, y_{2}=3$. Thus, we have two points of intersection: $A(-5 ; 45)$ and $B(2 ; 3)$. Next, we find the derivatives of the given functions $y^{\prime}=4 x$ and $y^{\prime}=2 x-3$, and then the slopes of the tangents to the given curves at point $A$: $k_{1}=4(-5)=-20, k_{2}=2(-5)-3=-13$. Therefore, the equations of the tangents at point $A$ are

$$
y-45=-20(x+5) \text { and } y-45=-13(x+5)
$$

or

$$
y=-20 x-55 \text { and } y=-13 x-20
$$

Similarly, for point $B$ we find $k_{1}=8$ and $k_{2}=1$, and the equations of the tangents are written as

$$
y=8 x-13 \text { and } y=x+1
$$