Odpowiedź :
A.
[tex]\Delta = b^2 - 4ac = 1^2 - 4 \cdot 2 \cdot \left( -1 \right) = 1 + 9 = 9\\x_1 = \frac{-b+\sqrt{\Delta} }{2a} = \frac{-1 + 3}{2 \cdot 2} = \frac{2}{4} = \frac{1}{2} \\x_2 = \frac{-b-\sqrt{\Delta} }{2a} = \frac{-1 - 3}{2 \cdot 2} = \frac{-4}{4} = -1\\x \in \langle -1;\ \frac{1}{2} \rangle[/tex]
B.
[tex]\Delta = 6^2 - 4 \cdot 1 \cdot 9 = 36 - 36 = 0\\x_0 = \frac{-6}{2} = -3\\x \in \mathbb{R} \setminus \{ -3 \}[/tex]
C.
[tex]-x^2 + 2x - 6 > 0\\\Delta = 2^2 - 4 \cdot (-1) \cdot (-6) = 4 - 24 = -20\\x \in \varnothing[/tex]
D.
[tex]\Delta = (-1)^2 - 4 \cdot 1 \cdot 1 = 1 - 4 = -3\\x \in \mathbb{R}[/tex]
E.
[tex](2x + 1)^2 + (x - 3)^2 = 4x^2 + 4x + 1 + x^2 - 6x + 9 = 5x^2 - 2x + 10\\5x^2 - 2x + 10 - 10 = 5x^2 - 2x\\\Delta = (-2)^2 - 4 \cdot 5 \cdot 0 = 4\\x_1 = \frac{2 - 2}{10} = 0\\x_2 = \frac{2 + 2}{10} = \frac{4}{10} = \frac{2}{5} \\x \in \left( 0;\ \frac{2}{5} \right)[/tex]