Question

Решите 
х^2+(x+4)=(18-(х+4) -х)^2

Answer 1

$x^2+(x+4)=(18-(x+4)-x)^2$
$x^2+x+4=(14-2x)^2$
$x^2+x+4=196-56x+4x^2$
$3x^2-57x+192=0$
$x^2-19x+64=0$
$D=(-19)^2-4*1*64=105$
$x_{1,2}=frac{19^+_-sqrt{105}}{2}$