А) 5х2 – 10 = 0; б) х2 + 4х = 0; в) 3х2 + 7х + 2 = 0; г) х2 – 8х + 12 = 0; д) х2 + х + 3...

Решение:
$5 {x}^{2} - 10 = 0 \\ 5 {x}^{2} = 10 \\ {x}^{2} = 2 \\ x = \sqrt{2}$

${x}^{2} + 4x = 0 \\ x(x + 4) = 0 \\ x_1 = 0 \\ x_2 = - 4$

$3 {x}^{2} + 7x + 2 = 0 \\ D = 49 - 4 \times 3 \times 2 = 49 - 24 = 25 \\ x_1 = \frac{ - 7 + 5}{6} = - \frac{1}{3} \\ x_2 = \frac{ - 7 - 5}{6} = - 2$

${x}^{2} - 8x + 12 = 0 \\ D_1 = 16 - 12 = 4 \\ x_1 = 4 + 2 = 6 \\ x_2 = 4 - 2 = 2$

${x}^{2} + x + 3 = 0 \\ D = 1 - 12 < 0$
Корней нет

$(2x - 1)(2x + 1) - (x - 3)(x + 1) = 18 \\ 4 {x}^{2} - 1 - {x}^{2} - x + 3x + 3 = 18 \\ 3 {x}^{2} + 2x - 16 = 0 \\ D_1 = 1 + 48 = 49 \\ x_1 = \frac{ - 1 + 7}{3} = 2 \\ x_2 = \frac{ - 1 - 7}{3} = - 2 \frac{2}{3}$
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Примечание:

$D=b^{2}-4ac\\x_{1,2}=\frac{-b\pm \sqrt{D}}{2a}\\\\D_1 = k^{2}-ac\\x_{1,2}=\frac{-k\pm \sqrt{D_1}}{a}\\k=\frac{1}{2}b$