(у+2)^2=49c^2=7

Решите задачу:

$A^2-B^2=(A-B)(A+B)\\\\1)\; \; (y+2)^2=49\\\\(y+2)^2-7^2=0\\\\(y+2-7)(y+2+7)=0\\\\(y-5)(y+9)=0\\\\y-5=0\; \; \; ili\; \; \; y+9=0\\\\y_1=5\; \; \; \; ili\; \; \; \; \; y=-9\\\\2)\; \; \frac{1}{4}c^2=7$

$(\frac{1}{2}c)^2-(\sqrt7)^2=0\\\\(\frac{1}{2}c-\sqrt7)(\frac{1}{2}c+\sqrt7)=0\\\\\frac{1}{2}c-\sqrt7=0\; \; \; ili\; \; \; \frac{1}{2}c+\sqrt7=0\\\\\frac{1}{2}c=\sqrt7\; \; \; \; ili\; \; \; \; \frac{1}{2}c=-\sqrt7\\\\c=2\sqrt7\; \; \; \; ili\; \; \; \; c=-2\sqrt7$