-14x² - 11x + 15 = 0 x1 = x2=

We have the quadratic equation $$-14x^2 - 11x + 15 = 0$$. We can multiply by -1 to make the leading coefficient positive: $$14x^2 + 11x - 15 = 0$$.

We can use the quadratic formula to solve for $$x$$:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

In this case, $$a = 14$$, $$b = 11$$, and $$c = -15$$.

$$x = \frac{-11 \pm \sqrt{11^2 - 4(14)(-15)}}{2(14)}$$

$$x = \frac{-11 \pm \sqrt{121 + 840}}{28}$$

$$x = \frac{-11 \pm \sqrt{961}}{28}$$

$$x = \frac{-11 \pm 31}{28}$$

We have two cases:

Case 1: $$x = \frac{-11 + 31}{28} = \frac{20}{28} = \frac{5}{7}$$

Case 2: $$x = \frac{-11 - 31}{28} = \frac{-42}{28} = -\frac{3}{2}$$

Thus, we have $$x_1 = \frac{5}{7}$$ and $$x_2 = -\frac{3}{2}$$.

Ответ: $$x_1 = \frac{5}{7}$$, $$x_2 = -\frac{3}{2}$$