Odpowiedź:
[tex](x+1)(x+1)-\frac{2x-6}{3}=(x-5)^2+\frac{3x+5}{2}-3\\\\x^2+x+x+1-\frac{2x-6}{3}=x^2-2\cdot x\cdot5+5^2+\frac{3x+5}{2}-3\\\\x^2+2x+1-\frac{2x-6}{3}=x^2-10x+25+\frac{3x+5}{2}-3\\\\x^2+2x+1-\frac{2x-6}{3}=x^2-10x+22+\frac{3x+5}{2} \ \ |\cdot6\\\\6x^2+12x+6-2(2x-6)=6x^2-60x+132+3(3x+5)\\\\6x^2+12x+6-4x+12=6x^2-60x+132+9x+15\\\\6x^2+8x+18=6x^2-51x+147\\\\6x^2+8x-6x^2+51x=147-18\\\\59x=129 \ \ |:59\\\\x=\frac{129}{59}\\\\\boxed{\boxed{x=2\frac{11}{59}}}[/tex]
