résoudre (3x 7)au carré = (7x-10) au carré

Bonjour Hanaiti

[tex](3x+7)^2=(7x-10)^2\\(3x+7)^2-(7x-10)^2=0\\\ [(3x+7)+(7x-10)][(3x+7)-(7x-10)]=0\\(3x+7+7x-10)(3x+7-7x+10)=0\\(10x-3)(-4x+17)=0\\10x-3=0\ \ ou\ \ -4x+17=0\\10x=3\ \ ou\ \ -4x=-17\\\\x=\dfrac{3}{10}\ \ ou\ \ x=\dfrac{-17}{-4}\\\\\boxed{x=\dfrac{3}{10}\ \ ou\ \ x=\dfrac{17}{4}}[/tex]