2x2+2x–1=0
D=22–4·2·(–1)=4+8=12
x1=(–2–2√3)/4; x2=(–2+2√3)/4;
x1=(–1–√3)/2; x2=(–1+√3)/2;
[m] S= ∫_{\frac{(-1-\sqrt{3}}{2}} ^{\frac{(-1+\sqrt{3}}{2}}(-x^2-3-(x^2+2x-4))dx= ∫_{\frac{(-1-\sqrt{3}}{2}} ^{\frac{(-1+\sqrt{3}}{2}}(1-2x-x^2)dx=[/m]
[m]=(x-x^2-\frac{x^3}{3})_{\frac{(-1-\sqrt{3}}{2}}| ^{\frac{(-1+\sqrt{3}}{2}}=(\frac{(-1+\sqrt{3}}{2}-(\frac{(-1+\sqrt{3}}{2})^2-\frac{(\frac{(-1+\sqrt{3}}{2})^3}{3})-(\frac{(-1-\sqrt{3}}{2}-(\frac{(-1-\sqrt{3}}{2})^2-\frac{(\frac{(-1-\sqrt{3}}{2})^3}{3})=...[/m]
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