1. [latex](x-5)(x+6)=-30 \ x^2+6x-5x-30=-30 \ x^2+x=0 \ x(x+1)=0 \ x=0 ext{ lub }x=-1[/latex] 2. Liczba parzysta - [latex]2n[/latex] [latex](2n)^2+(2n+2)^2=244 \ 4n^2+4n^2+8n+4=244 \ 8n^2+8n-240=0 |:8\ n^2+n-30=0 \ Delta=1^2-4*1*(-30)=121 \ sqrt{Delta}=11 \ n_1= frac{-1-11}{2} = frac{ -12}{2}=-6 otin N\ n_2= frac{-1+11}{2}=5 in N \ (2*5)^2+(2*5+2)^2=100+144=244 \[/latex] Zatem rozwiązaniem są liczby 10 i 12
[latex]1.\(x -5)(x+6) = -30\\ x^{2} + 6x - 5x - 30 + 30 = 0\\ x^{2} + x = 0\\x(x + 1) = 0\\x = 0 V x = -1[/latex] [latex]2)\2n - I liczba\2n+2 II liczba\\(2n)^{2} + (2n+2)^{2} = 244\\4n^{2} + 4n^{2} + 8n + 4 - 144 = 0\\8n^{2} + 8n -240 = 0 /:8\\n^{2} + n - 30 = 0\\Delta = b^2 - 4ac = 1 + 120 = 121\\sqrt{Delta} = 11\ _1 = frac{-1-11}{2} = -6 otin N\\n_2 = frac{-1+11}{2} = 5[/latex] [latex]n = 5\\I liczba = 2n = 2 * 5 = 10\\II liczba = 2n + 2 = 2 *5 +2 = 12\\Spr.\10^{2} + 12^{2} = 100 + 144 = 244[/latex]
