Odpowiedź:
[tex]Zad.20\\\\a)\\\\(x+3)(x+5)=-4(x^2+2)+5x^2\\\\x^2+5x+3x+15=-4x^2-8+5x^2\\\\x^2+8x+15=x^2-8\\\\x^2+8x-x^2=-8-15\\\\8x=-23\ \ /:8\\\\x=-\frac{23}{8}\\\\x=-2\frac{7}{8}\\\\\\b)\\\\(x-2)(4+x)=x(x-5)\\\\4x+x^2-8-2x=x^2-5x\\\\2x+x^2-8=x^2-5x\\\\2x+x^2-x^2+5x=8\\\\7x=8\ \ /:7\\\\x=\frac{8}{7}\\\\x=1\frac{1}{7}[/tex]
[tex]c)\\\\-2x(x+1)=(3-2x)(x+2)\\\\-2x^2-2x=3x+6-2x^2-4x\\\\-2x^2-2x-3x+2x^2+4x=6\\\\-5x+4x=6\\\\-x=6\ \ /\cdot(-1)\\\\x=-6\\\\\\d)\\\\(x-3)(x+3)=(x-2)^2\\\\x^2-9=x^2-4x+4\\\\x^2-x^2+4x=4+9\\\\4x=13\ \ /:4\\\\x=\frac{13}{4}\\\\x=3\frac{1}{4}[/tex]
