6x+|4-x^2|=3найти сумму корней уравнения

$6x+|4-x^2|=3$
1) $\left \{ {{4-x^{2} \geq 0} \atop {6x+4-x^2=3}} \right.$
$\left \{ {{-2 \leq x \leq 2} \atop {x^2-6x-1=0, D=40}} \right.$
$\left \{ {{-2 \leq x \leq 2} \atop {x_{1}= \frac{6- \sqrt{40}}{2}, x_{2}= \frac{6+\sqrt{40}}{2}}} \right.$
$\left \{ {{-2 \leq x \leq 2} \atop {x_{1}=3- \sqrt{10}, x_{2}=3+\sqrt{10}} \right.$
$x= 3-\sqrt{10}$ - корень

2) $\left \{ {{4-x^{2}\leq0} \atop {6x-4+x^2=3}} \right.$
$\left \{ {{x \leq -2, x \geq 2} \atop {x^2+6x-7=0, D=64}} \right.$
$\left \{ {{x \leq -2, x \geq 2} \atop {x_{3}=-7, x_{4}=1}} \right.$
$x=-7$ - корень

Сумма корней: $3-\sqrt{10}-7=-4-\sqrt{10}$