Задание 266

$$x^2+2x=3\cdot (4-x^2)$$

$$x^2+2x=12-3x^2$$

$$x^2+3x^2+2x-12=0$$

$$4x^2+2x-12=0$$

$$2x^2+x-6=0$$

$D=1^2-4\cdot 2\cdot (-6)=$$1+48=49$

$x_1=\frac{-1+\sqrt{49}}{2\cdot 2}=$$\frac{-1+7}{4}=\frac{6}{4}=1,5$

$$x_2=\frac{-1-7}{4}=\frac{-8}{4}=-2$$