8 класс решите тут 2 примера

$(2x+1)^2=(1+3x)(3x-1)\\ (2x+1)^2=(3x+1)(3x-1)\\ 4x^2+4x+1=9x^2-1\\ 5x^2-4x-2=0;\ D=(-4)^2-4*5*(-2)=16+40=56=(2 \sqrt{14})^2 \\ x_{1,2}= \frac{-(-4)\pm 2 \sqrt{14} }{2*5} = \frac{2\pm \sqrt{14} }{5}$
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$\frac{2x^2+3x}{5} + \frac{3x^2+4x}{7} = \frac{5-x}{2}\\\\ \frac{2*7*(2x^2+3x)}{2*7*5} + \frac{2*5*(3x^2+4x)}{2*5*7} - \frac{5*7*(5-x)}{5*7*2}=0\\\\ \frac{14*(2x^2+3x)+10*(3x^2+4x)-35*(5-x)}{5*7*2}=0\\\\ \left \{ {14*2x^2+14*3x+10*3x^2+10*4x-35*5+35*x=0} \atop {5*7*2 \neq 0}} \right. \\\\ \left \{ {28x^2+42x+30x^2+40x-175+35x=0} \atop {5*7*2 \neq 0}} \right. \\\\ 58x^2+117x-175=0\\\\ D=117^2-4*58*(-175)=13689+40600=54289=233^2\\\\ x_{1,2}= \frac{-117\pm233}{2*58} =- \frac{175}{58}\ or\ 1$