Odpowiedź :
Rozwiązanie:
b)
[tex]x(x-5)=2x+10\\x^{2}-5x=2x+10\\x^{2}-7x-10=0\\\Delta=49-4*1*(-10)=89\\x_{1}=\frac{7-\sqrt{89} }{2} \\x_{2}=\frac{7+\sqrt{89} }{2}[/tex]
c)
[tex](x-3)^{2}=9-4x\\x^{2}-6x+9=9-4x\\x^{2}-2x=0\\x(x-2)=0\\x=0, x=2[/tex]
B) x(x-5)=2x+10
X^2-5x-2x=10
x^2-7x=10
x(x-7)=10
x=10 lub x-7=10
x=17
c) (x-3)^2=9-4x
x^2-6x+9=9-4x
x^2-6x+4x=9-9
x^2-2x= 0
x(x-2)= 0
x= 0 lub x-2= 0
x= 2