Question

Х(х-3)=28 и х(х+1)=42

Answer 1

$x^{2} -3X-28=0$
$\sqrt{D} =9-4*-28=9+112=121$
D=11
X1=(3-11)/2=-4
X2=(3+11)/2=7
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$x^{2} +x-42=0$
$\sqrt{D} =1-4*-42=1+168=169$
D=13
x1=(-1-13)/2=-7
x2=(-1+13)/2=6