Question

(x+5)^2-(2-x)^2=0 решить уравнение

Answer 1

$(x+5)^2-(2-x)^2=0\\x^2+10x+25-(4-4x+x^2)=0\\x^2+10x+25-4+4x-x^2=0\\14x+21=0\\14x=-21\\x=-\frac{3}{2} =-1,5$

Answer 2

$\displaystyle \tt (x+5)^2-(2-x)^2=0\\\displaystyle \tt x^2+10x+25-(4-4x+x^2)=0\\\displaystyle \tt x^2+10x+25-4+4x-x^2=0\\\displaystyle \tt 14x+21=0\\\displaystyle \tt 14x=-21\\\displaystyle \tt x=-21\div14\\\displaystyle \tt \bold{x=-1,5}$