Rozwiązane

rozwiąż równania
2x^{2} +8x -10 = 0
-5x - 4x^{2} = 0



Odpowiedź :

1)

[tex]2x^2+8x-10=0 \ \ /:2\\\\x^2+4x-5=0\\\\a=1, \ b=4, \ c=-5\\\\\Delta=b^2-4ac\Rightarrow4^2-4\cdot1\cdot(-5)=16+20=36\\\\\sqrt{\Delta}=\sqrt{36}=6\\\\x_1=\frac{-b-\sqrt{\Delta}}{2a}\Rightarrow\frac{-4-6}{2\cdot1}=\frac{-10}{2}=\boxed{-5}\\\\x_2=\frac{-b+\sqrt{\Delta}}{2a}\Rightarrow\frac{-4+6}{2\cdot1}=\frac{2}{2}=\boxed1[/tex]

2)

[tex]-5x-4x^2=0 \ \ /\cdot(-1)\\\\4x^2+5x=0\\\\4x(x+1,25)=0\\\\4x=0\Rightarrow\boxed{x=0}\\\\x+1,25=0\Rightarrow\boxed{x=-1,25}[/tex]

Odpowiedź:

2x² + 8x - 10 = 0

a =2 , b = 8 , c = -  10

Δ = b² - 4ac = 8² - 4 * 2 * (- 10) = 64 + 80 = 144

√Δ = √144 = 12

x₁ = (- b - √Δ)/2a = (- 8 -  12)/4 = - 20/4 = - 5

x₂ = (- b + √Δ)/2a = (-  8 + 10)/4 = 2/4 = 1/2

- 5x + 4x² = 0

x(- 5 + 4x) =  0

x = 0 ∨ - 5 + 4x = 0

x = 0 ∨ 4x = 5

x = 0 ∨ x =5/4 = 1 1/4  

∨ - znaczy "lub"