Odpowiedź:
[tex](x-4)(2x+5)=(x-4)(2-x)\\\\2x^2+5x-8x-20=2x-x^2-8+4x\\\\2x^2-3x-20=6x-x^2-8\\\\2x^2-3x-20-6x+x^2+8=0\\\\3x^2-9x-12=0\ \ /:3\\\\x^2-3x-4=0\\\\\Delta=b^2-4ac=(-3)^2-4*1*(-4)=9+16=25\\\\\sqrt{\Delta}=\sqrt{25}=5\\\\\\x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-5}{2*1}=\frac{3-5}{2}=\frac{-2}{2}=-1\\\\\\x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+5}{2*1}=\frac{3+5}{2}=\frac{8}{2}=4[/tex]
