(x+3)(x+7)-(x+4)^2=0 (y+8)^2-(y+9)(y-5)=117

Решите задачу:

$1)(x+3)(x+7)-(x+4) ^{2} =0 \\ x^{2} +3x+7x+21- (x^{2} +16+8x)=0\\ x^{2} +10x+21- x^{2} -16-8x=0 \\ 2x+5=0 \\ 2x=-5 \\ x=- \frac{5}{2} \\ x=-2.5 \\ \\ \\$

$2)(y+8) ^{2} -(y+9)(y-5)=117\\y ^{2} +64+16y-(y ^{2} -5y+9y-45)=117 \\ y ^{2} +64+16y-(y ^{2} +4y-45)=117\\y ^{2} +64+16y-y ^{2} -4y+45=117\\16y-4y=117-64-45\\12y=8 \\ y= \frac{8}{12} \\ y= \frac{2}{3}$