2x+√3x-2=3 Найти решение

$2x+\sqrt{3x-2}=3$

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ОДЗ: $3x-2\geq 0$

$3x\geq 2$

$x\geq \frac{2}{3}$

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$\sqrt{3x-2}=3-2x \\ \\ \left \{ {{3-2x\geq0} \atop {(\sqrt{3x-2})^2}=(3-2x)^2} \right. \\ \\ \left \{ {{-2x\geq -3} \atop {3x-2=3^2-2\cdot 3\cdot 2x+(2x)^2} \right. \\ \\ \left \{ {{x\leq 1,5} \atop {3x-2=9-12x+4x^2} \right. \\ \\ \left \{ {{x\leq 1,5} \atop {4x^2-15x+11=0} \right. \\ \\ 4x^2-15x+11=0 \\ \\ D=(-15)^2-4\cdot 4 \cdot 11=225-176=49 \\ \\ x_{1,2}=\frac{-(-15)\pm\sqrt{49}}{2\cdot4}=\frac{15\pm7}{8} \\ \\ \left \{ {{x\leq 1,5} \atop {x_1=1, x_2=\frac{11}{4}} \right. \\ \\ x=1$

Ответ: 1