Odpowiedź :
f(x)=4x²-28x+33
4x²-28x+33=0
Δ=(-28)²-4·4·33
Δ=784-528
Δ=256
√Δ=√256=16
x₁=(28-16)/(2·4)=12/8=1 1/2
x₂=(28+16)/(2·4)=44/8=5 1/2
Funkcja ma dwa miejsca zerowe x₁=1 1/2 i x₂=5 1/2
[tex]f(x)=4x^2-28x+33\\\\4x^2-28x+33=0\\\\a=4, \ b=-28, \ c=33\\\\\Delta=b^2-4ac\rightarrow(-28)^2-4\cdot4\cdot33=784-528=256\\\\\sqrt{\Delta}=\sqrt{256}=16\\\\x_1=\frac{-b-\sqrt{\Delta}}{2a}\rightarrow\frac{-(-28)-16}{2\cdot4}=\frac{12}{8}=\boxed{1,5}\\\\x_2=\frac{-b+\sqrt{\Delta}}{2a}\rightarrow\frac{-(-28)+16}{2\cdot4}=\frac{44}{8}=\boxed{5,5}[/tex]