1,
Áp dụng phương trình hoành độ giao điểm, ta có:
$-2x²=x-3$
$⇔2x²+x-3=0$
$⇔(2x+3)(x-1)=0$
$⇔\left[ \begin{array}{l}x=1⇒y=-2\\x=\dfrac{-3}{2}⇒y=\dfrac{-7}{2}\end{array} \right.$
2,
a, $3x²+2x-1=0$
$⇔(x+1)(3x-1)=0$
$⇔\left[ \begin{array}{l}x=\dfrac{1}{3}\\x=-1\end{array} \right.$
b, $2x²-3x-5=0$
$⇔(x+1)(2x-5)=0$
$⇔\left[ \begin{array}{l}x=\dfrac{5}{2}\\x=-1\end{array} \right.$