√2х+5-√х+6=1 помагите

$\sqrt{2x+5}- \sqrt{x+6}=1;$
ОДЗ: $\left \{ {{2x+5 \geq 0} \atop {x+6 \geq 0}} \right. \Rightarrow \left \{ {{x\geq-2,5} \atop {x \geq-6}} \right. \Rightarrow x\geq-2,5$
$(\sqrt{2x+5}- \sqrt{x+6})^2=1^2;$
$2x+5-2\sqrt{2x+5} \sqrt{x+6}+x+6=1;$
$2\sqrt{2x+5}\sqrt{x+6}=3x+10;$
$(2\sqrt{(2x+5)(x+6)})^2=(3x+10)^2;$
$4(2x+5)(x+6) =9 x^{2}+ 60x+100;$
$8x^{2}+68x+120- 9 x^{2}- 60x-100=0;$
$- x^{2}+8x+20=0;x^{2}-8x-20=0;$
$D_1=16+20=36;x_1=-2;x_2=10;$
Ответ: $x=-2;x=10;$