Помогите,срочно! √(2x + 3) + √(4 - x) = √(3x + 7)

$\sqrt{2x+3} + \sqrt{4-x} = \sqrt{3x+7}$

ОДЗ:

$\left \{ {{2x+3 \geq 0} \atop {4-x \geq 0}} \atop {3x+7 \geq 0}} \right.$

$\left \{ {{x \geq -1.5} \atop {x \leq 4}} \atop {x \geq -2 \frac{1}{3} }} \right.$

$x$ ∈ $[-1.5;4]$

$(\sqrt{2x+3} + \sqrt{4-x} )^2= (\sqrt{3x+7} )^2$

$2x+3+4-x+2 \sqrt{(2x+3)(4-x}) =3x+7$

$x+7+2 \sqrt{(2x+3)(4-x}) =3x+7$

$2 \sqrt{(2x+3)(4-x}) =2x$

$\sqrt{(2x+3)(4-x}) =x$

$\left \{ {{x \geq 0} \atop {(2x+3)(4-x) =x^2}} \right.$

$8x-2x^2+12-3x =x^2$

$3x^2-5x-12=0$

$D=(5)^2-4*3*(-12)=25+144=169=13^2$

$x_1= \frac{5+13}{6}=3$

$x_2= \frac{5-13}{6}=- \frac{8}{6} =- \frac{4}{3} =-1 \frac{1}{3}$ ∅

Ответ: $3$