`2x-2x^2+7=0`
`<=>2x^3-2x-7=0`
`<=>x^2-x-7/2=0`
`<=>x^2-2.x. 1/2+1/4-15/4=0`
`<=>(x-1/2)^3=15/4`
`<=>x=(+-sqrt15+1)/2`
Ta có: \(-2x^2+2x+7=0\)
a=-2; b=2; c=7
\(\text{Δ}=2^2-4\cdot\left(-2\right)\cdot7=4+56=60>0\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{-b-\sqrt{\text{Δ}}}{2a}\\x_2=\dfrac{-b+\sqrt{\text{Δ}}}{2a}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_1=\dfrac{-2-2\sqrt{15}}{-4}=\dfrac{1+\sqrt{15}}{2}\\x_2=\dfrac{-2+2\sqrt{15}}{-4}=\dfrac{1-\sqrt{15}}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{1+\sqrt{15}}{2};\dfrac{1-\sqrt{15}}{2}\right\}\)
