Задание в картинках...

Решите задачу:

$x^{2}+2x=m\\\\2m-\frac{3}{m}-5=0\\\\\frac{2m^{2}-5m-3 }{m}=0,m\neq0\\\\2m^{2}-5m-3=0\\\\D=(-5)^{2}-4*2*(-3)=25+24=49=7^{2}\\\\m_{1}=\frac{5+7}{4}=3\\\\m_{2}=\frac{5-7}{4}=-0,5$

$1)x^{2}+2x=-0,5\\\\x^{2}+2x+0,5=0\\\\D=2^{2}-4*(0,5)=4-2=2\\\\x_{1}=\frac{-2+\sqrt{2} }{2}\\\\x_{2}=\frac{-2-\sqrt{2} }{2}$

$x^{2}+2x=3\\\\x^{2}+2x-3=0$

$D=2^{2}-4*(-3)=4+12=16=4^{2}\\\\x_{3}=\frac{-2+4}{2} =1\\\\x_{4}=\frac{-2-4}{2} =-3$