Verified answer

[tex]\displaystyle 1)x^2+7=8x\\x^2-8x+7=0\\D=(-8)^2-4*1*7=36\\\\x=\frac{8б\sqrt{36} }{2}=\frac{8б6}{2}\\ \\x_1=1,x_2=7\\ \\2)x^2-36=0\\D=0^2-4*1(-36)=144\\\\x=\frac{б\sqrt{144} }{2}=\frac{б12}{2}\\ \\x_1=-6,x_2=6\\ \\3)x^2=-15x\\x^2+15x=0\\D=15^2-4*1*0=225\\\\x=\frac{-15б\sqrt{225} }{2}=\frac{-15б15}{2}\\ \\x_1=-15,x_2=0[/tex]