(x + 3)(x − 4) = −12
$x^2 + 3x - 4x - 12 = -12$
$x^2 - x = -12 + 12$
$x^2 - x = 0$
x(x − 1) = 0
x = 0 или x − 1 = 0
x = 1
Ответ:
$x_1 = 0$;
$x_2 = 1$.

$1\frac{2}{3}t + (2t + 1)(\frac{1}{3}t - 1) = 0$
$1\frac{2}{3}t + \frac{2}{3}t^2 + \frac{1}{3}t - 2t - 1 = 0$
$1\frac{2}{3}t + \frac{2}{3}t^2 - 1\frac{2}{3}t - 1 = 0$
$\frac{2}{3}t^2 - 1 = 0$
$\frac{2}{3}t^2 = 1$
$t^2 = \frac{3}{2}$
$t = ±\sqrt{1,5}$

3x(2x + 3) = 2x(x + 4,5) + 2
$6x^2 + 9x = 2x^2 + 9x + 2$
$6x^2 - 2x^2 + 9x - 9x = 2$
$4x^2 = 2$
$x^2 = \frac{1}{2}$
$x = ±\frac{1}{\sqrt{2}}$

$(x - 1)(x + 1) = 2(x^2 - 3)$
$x^2 - x + x - 1 = 2x^2 - 6$
$x^2 - 2x^2 = -6 + 1$
$-x^2 = -5$
$x^2 = 5$
$x = ±\sqrt{5}$