rozwiąż równanie : (2x-2)do kwadratu - (2x+3)do kwadratu = -3(x-1)(x+1)   rozwiąż nierówność a)y+2/3- y-2/2>/ 1- y b) (8z+3)(8z-3)+(6z+2)do kwadratu < (10z+1)(10z-1)+4(2z+7)

[latex](2x-2)^{2}-(2x+3)^{2}=-3(x-1)(x+1)[/latex] [latex]4x^{2}-8x+4-(4x^{2}+12x+9)=-3(x^{2}-1)[/latex] [latex]4x^{2-8x+4-4x^{2}-12x-9=-3x^{2}+3[/latex] [latex]3x^{2}-20x-8=0[/latex] delta=400-4*3*(-8) delta=400+96 delta=496 pierwiastek z delty=[latex]4sqrt{31}[/latex] [latex]x_1=frac{20-4sqrt{31}}{6}[/latex] [latex]x_2=frac{20+4sqrt{31}}{6}[/latex]   b) (8z+3)(8z-3)+(6z+2)do kwadratu < (10z+1)(10z-1)+4(2z+7) [latex](64z^{2}-9)(36z^{2}+24z+4)<100z^{2}-1+8z+28[/latex] [latex]2304z^{4}+1536z^{3}+256z^{2}-324z^{2}-216z-36-100z^{2}-8z-27<0[/latex] [latex]2304z^{4}+1536z^{3}-168z^{2}-224z-63<0[/latex]