X^2/x+3=1/4 срочнооооо

[latex] \frac{x^2}{x+3}= \frac{1}{4} \\ (x+3)=4x^2\\4x^2-x-3=0\\ D=b^2-4ac=(-1)^2-4*4*(-3)=1+48=49\\ x_1= \frac{-b+ \sqrt{D} }{2a}= \frac{1+ \sqrt{49} }{8} = \frac{1+7}{8} =1\\ x_2= \frac{-b- \sqrt{D} }{2a} = \frac{1- \sqrt{49} }{8} = \frac{1-7}{8} = \frac{-6}{8}=-0.75[/latex] Ответ: [latex]1;-0.75[/latex]