1. a)
[tex] - {x}^{2} + 2x + 3 > 0[/tex]
[tex]∆ = {b}^{2} - 4ac = {2}^{2} - 4 \times ( - 1) \times 3 = 4 + 12 = 16[/tex]
[tex]√∆ = 4[/tex]
[tex] x_{1} = \frac{ - b - √∆}{2a} = \frac{ - 2 - 4}{2 \times ( - 1)} = \frac{ - 6}{ - 2} = 3[/tex]
[tex] x_{2} = \frac{ - b + √∆}{2a} = \frac{ - 2 + 4}{2 \times ( - 1)} = \frac{2}{ - 2} = - 1[/tex]
[tex]x€( - 1;3)[/tex]
b)
[tex] {2x}^{2} - 6x > 0[/tex]
[tex]2x(x - 3) > 0[/tex]
[tex]2x = 0 | \div 2| [/tex]
[tex]x = 0[/tex]
[tex]x - 3 = 0[/tex]
[tex]x = 3[/tex]
[tex]x€(0;3)[/tex]
2.
[tex] {4x}^{2} - {(x + 2)}^{2} = 0[/tex]
[tex] {4x}^{2} - ( {x}^{2} + 2 \times x \times 2 + {2}^{2}) = 0[/tex]
[tex] {4x}^{2} - {x}^{2} - 4x - 4 = 0[/tex]
[tex] {3x}^{2} - 4x - 4 = 0[/tex]
[tex]∆ = {( - 4)}^{2} - 4 \times 3 \times ( - 4) = 16 + 48 = 64[/tex]
[tex]√∆ = 8[/tex]
[tex] x_{1} = \frac{ - ( - 4) - 8}{2 \times 3} = \frac{4 - 8}{6} = \frac{ - 4}{6} = - \frac{2}{3} [/tex]
[tex] x_{2} = \frac{ - ( - 4) + 8}{2 \times 3} = \frac{4 + 8}{6} = \frac{12}{6} = 2[/tex]
