A-b=10 a*b=7 (a+b)^2=?

Составим систему:
$\left \{ {{a-b=10} \atop {a*b=7}} \right. \Rightarrow \left \{ {{a=10+b} \atop {b(10+b)=7}} \right. \Rightarrow \left \{ {{a=10+b} \atop {10b+b^2=7}} \right.$

Решим:

$b^2+10b-7=0$
$\sqrt{D}= \sqrt{100+28}= \sqrt{128}=4\sqrt{8}=8\sqrt{2}$
$b_{1,2}= \frac{-10\pm8 \sqrt{2}}{2}=- \frac{2(5\pm4 \sqrt{2})}{2}=-5\pm4 \sqrt{2}$

Найдем a:
$a_1=10-5+4\sqrt{2} =5+4\sqrt{2}=5+4\sqrt{2}$
$a_2=10-5-4\sqrt{2}=5-4 \sqrt{2}=5-4\sqrt{2}$

Отсюда получаем:
$(a_1+b_1)^2=[(5+4 \sqrt{2})+(-5+4 \sqrt{2})]^2=(8 \sqrt{2})^2=64*2=128$
$(a_2+b_2)^2=[(5-4 \sqrt{2})+(-5-4 \sqrt{2})]^2=(-8 \sqrt{2})^2=64*2=128$

Отсюда следует:
$(a+b)^2=128$