A not so hard basic calculus problem? But it appears to be very lengthy Find the coordinates of the two points on the curve $y=4-x^2$ whose tangents pass through the point $(-1,7)$.
My work:
Let the two points be $(a,b)$ and $(c,d)$. And $\frac{dy}{dx}=-2x$, so the gradients of the tangents are $-2a$ and $-2c$. We get $\frac{7-d}{-1-c}=-2c\implies 7-d=2c+2c^2$ and $\frac{7-b}{-1-a}=-2a\implies 7-b=2a+2a^2$. Now we have 2 equations and 4 unknowns. So we need two more equations.
We have the equations of the tangents are $y=-2cx+2c^2+d$ and $y=-2ax+2a^2+b$. The solutions of these two equations are $(-1,7)$, putting $(-1,7)$ in we get $7(a-c)=-2ac(a-c)+ad-bc$ and $2(a-c)=-2(a-c)(a+c)+d-b$.
We can also equate $4-x^2=-2cx+2c^2+d$ and $4-x^2=-2ax+2a^2+b$, then we get $a^2-c^2=d-b$. From here, we still have to substitute and do some algebra to get $(a,b)$ and $(c,d)$.
Is my working correct? But I feel like the working is too lengthy. Is there any faster way?
Many thanks!

Your work is correct. Maybe you can shorten it by assuming that the points of tangency is $P=(t,4-t^2)$. Then slope at $P$ is $m_P=-2t$. Slope of line $LP=\dfrac{7-(4-t^2)}{-1-t}=\dfrac{3+t^2}{-(1+t)}$. Now this must be equal to $m_P$.
$$\dfrac{3+t^2}{-(1+t)}=-2t.$$
Solving this gives
$$t^2+2t-3=0.$$
Now solve for the two values of $t$. 
