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

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