Помогите пожалуйста! Фото ниже.

7.
1.)
$a.\\(2x-5)^2-3x^2=(x+2)^2-3\\4x^2-20x+25-3x^2=x^2+4x+4-3\\x^2-20x+25=x^2+4x+1\\x^2-x^2+25-1=20x+4x\\24=24x\\24x=24\\x=\frac{24}{24}\\x=1\\Proverka:\\(2*1-5)^2-3*1^2=(1+2)^2-3\\(-3)^2-3=3^2-3\\9-3=9-3\\6=6$
$b.\\(2x-1)(x+1)-x^2=(x-3)^2-10\\2x^2+2x-x-1-x^2=x^2-6x+9-10\\x^2+x-1=x^2-6x-1\\x^2+x-1-x^2+6x+1=0\\7x=0\\x=0\\Proverka:\\(2*0-1)(0+1)-0^2=(0-3)^2-10\\(-1)(1)-0=(-3)^2-10\\-1-0=9-10\\-1=-1$
2)
$a.\\(y+1)^2-4y=-5+(y-2)^2\\y^2+2y+1-4y=-5+y^2-4y+4\\y^2-2y+1=y^2-4y-1\\y^2-y^2+4y-2y=-1-1\\2y=-2\\y=\frac{-2}{2}\\y=-1\\Proverka:\\(-1+1)^2-4*(-1)=-5+(-1-2)^2\\0^2+4=-5+(-3)^2\\4=-5+9\\4=4$
$b.\\(3x+1)(x-1)-2x^2=(x+2)^2-5\\3x^2-3x+x-1-2x^2=x^2+4x+4-5\\x^2-2x-1=x^2+4x-1\\x^2-x^2=4x+2x-1+1\\0=6xd\\0=x\\x=0\\Proverka:\\(3*0+1)(0-1)-2*0^2=(0+2)^2-5\\1*(-1)-0=2^2-5\\-1=4-5\\-1=-1$