Решительно систему уравнений (x-2)*(y+3)=160, y-x=1

$\left \{ {{(x-2)(y+3)=160} \atop {y-x=1}} \right.$
$\left \{ {{(x-2)(y+3)=160} \atop {y=1+x}} \right.$
$\left \{ {{(x-2)(1+x+3)=160} \atop {y=1+x}} \right.$
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(x-2)(1+x+3)=160
(x-2)(x+4)=160
$x^{2}$ +4x-2x-8=160
$x^{2}$ +2x-168=0
D=4+4*168=676
x1=12
x2= -14
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$\left \{ {{x1=12} \atop {y=1+12}} \right.$
$\left \{ {{x2=-14} \atop {y=1-14}} \right.$
$\left \{ {{x=12} \atop {y=-13}} \right.$