Simultaneous Quadratic Equations: $x^2 + y ^ 2 - 2 x + 6y - 35 = 0$ and $2x + 3y = 5$ I've been given the task to simultaneously solve:
$$x^2 + y ^ 2 - 2 x + 6y - 35 = 0$$
$$2x + 3y = 5$$
I've tried applying the substitution method by reordering the second equation to both $x$ and $y$, but both times I have got a surd.
According to the answers I should be getting $(-2, 3)$ and $(\frac{100}{13}, -\frac{45}{13})$, so evidently I am doing something wrong.
This is in the chapter Equations Reducible to Quadratics and I can not see how the material in the intro applies to the question.

From the second equation, we have that $x=\frac{5-3y}{2}$.  Plugging this into the first equation, 
$$\left(\frac{5-3y}{2}\right)^2+y^2-(5-3y)+6y-35=0\\
\frac{9y^2-30y+25}{4}+y^2+9y-40=0$$
Multiplying by $4$,
$$9y^2-30y+25+4y^2+36y-160=0\\
13y^2+6y-135=0$$
Applying the quadratic formula,
$$y=\frac{-6\pm\sqrt{6^2-4(13)(-135)}}{26}\\
=\frac{-6\pm84}{26}=\frac{-3\pm42}{13}=3,-\frac{45}{13}$$
Plugging these values back into our equation $x=\frac{5-3y}{2}$, we get solutions of $(2,-3)$, and $(\frac{100}{13},-\frac{45}{13}).$
