Cześć ancymonku ;-))
a)
[tex]2x^2+5x-12=0\\\\a=2, \ b=5, \ c=-12\\\\\Delta=5^2-4\cdot2\cdot(-12)=25+96=121\\\\\sqrt{\Delta}=\sqrt{121}=11\\\\x_1=\frac{-5-11}{2\cdot2}=\frac{-16}{4}=-4\\\\x_2=\frac{-5+11}{2\cdot2}=\frac{6}{4}=1,5[/tex]
d)
[tex]2(x^2+4)=-3x\\\\2x^2+8=-3x\\\\2x^2+3x+8=0\\\\a=2, \ b=3, \ c=8\\\\\Delta=3^2-4\cdot2\cdot8=9-64=-55[/tex]
Δ < 0, równanie nie posiada rozwiązania
b)
[tex]3x^2-7x=20\\\\3x^2-7x-20=0\\\\a=3, \ b=-7, \ c=-20\\\\\Delta=(-7)^2-4\cdot3\cdot(-20)=49+240=289\\\\\sqrt{\Delta}=\sqrt{289}=17\\\\x_1=\frac{-(-7)-17}{2\cdot3}=\frac{7-17}{6}=\frac{-10}{6}=-\frac{5}{3}\\\\x_2=\frac{-(-7)+17}{2\cdot3}=\frac{7+17}{6}=\frac{24}{6}=4[/tex]
e)
[tex]49x^2=4x\\\\49x^2-4x=0\\\\49x(x-\frac{4}{49})=0\\\\49x=0 \ \ \ \vee \ \ \ x-\frac{4}{49}=0\\\\x=0 \ \ \ \vee \ \ x=\frac{4}{49}[/tex]
c)
[tex]9x^2=12x-4\\\\9x^2-12x+4=0\\\\a=9, \ b=-12, \ c=4\\\\\Delta=(-12)^2-4\cdot9\cdot4=144-144=0\\\\x_0=\frac{-(-12)}{2\cdot9}=\frac{12}{18}=\frac{2}{3}[/tex]
f)
[tex]16x^2+25=40x\\\\16x^2-40x+25=0\\\\a=16, \ b=-40, \ c=25\\\\\Delta=(-40)^2-4\cdot16\cdot25=1600-1600=0\\\\x_0=\frac{-(-40)}{2\cdot16}=\frac{40}{32}=1,25[/tex]
Wykorzystane wzory
[tex]y=ax^2+bx+c\\\\\Delta=b^2-4ac\\\\x_1=\frac{-b-\sqrt{\Delta}}{2a} \ oraz \ x_2=\frac{-b+\sqrt{\Delta}}{2a} \ (gdy \ \Delta>0)\\\\x_0=\frac{-b}{2a} \ (gdy \ \Delta=0)[/tex]
