Корень (х+3) - корень(7-х)=2

Решите задачу:

$\sqrt{x+3} - \sqrt{7-x} =2 \\ ( \sqrt{x+3} )^2 =(2+ \sqrt{7-x})^2 \\ x+3=4+4\sqrt{7-x}+7-x \\ x+x-4-7+3=4\sqrt{7-x} \\ 2x-8=4\sqrt{7-x} \\ (2x-8)^2=(4\sqrt{7-x})^2 \\ 4x^2-32x+64=16(7-x) \\ 4x^2-32x+64=112-16x\\ 4 x^{2} -32x+16x+64-112=0 \\ 4 x^{2} -16x-48=0 \\ x^{2} -4x-12=0$
$D=(-4)^2-4*(-12)=16+48=64=8^2 \\ x_1= \frac{-(-4)- \sqrt{64} }{2*1} = \frac{4-8}{2}= \frac{-4}{2} =-2 \\ x_2= \frac{-(-4)+ \sqrt{64} }{2*1} = \frac{4+8}{2}= \frac{12}{2} =6 \\ \\ \\ \sqrt{(-2)+3} - \sqrt{7-(-2)} = \sqrt{1} - \sqrt{9} =1-3=-2 \\ -2 \neq 2 \\ \sqrt{6+3} - \sqrt{7-6} = \sqrt{9} - \sqrt{1} =3-1=2 \\ \\ Otvet: x=6$