PT này vô nghiệm.
Ta sẽ chứng minh $10x^2+24y^2+8x+20y+51 > 0 \forall
x,y.$
Thật vậy,
$10x^2+24y^2+8x+20y+51$
$=\left ( \sqrt{10}x \right )^2 + 2.
\sqrt{10}x.\frac{4}{ \sqrt{10}}+\frac{16}{10}+\left ( \sqrt{24}y \right
)^2 + 2. \sqrt{24}y .\frac{10}{ \sqrt{24}}+\frac{100}{24}+\frac{1357}{30}$
$=\left ( \sqrt{10}x+\frac{4}{ \sqrt{10}} \right )^2+\left (\sqrt{24}y+\frac{10}{ \sqrt{24}} \right )^2+\frac{1357}{30} > 0 \forall
x,y.$