Задание 1420

$$-\dfrac{13}{48} x^2=-9\dfrac34$$

$$x^2=-9\dfrac34:\left(-\dfrac{13}{48} \right)$$

$$x^2=\dfrac{39}4 \cdot \dfrac{48}{13}$$

$$x^2=3\cdot 12$$

$$x^2=36$$

$$x=\pm \sqrt{36}=\pm 6$$

$$6>-6$$