[tex]ax^2+bx+c=0\\\\\Delta=b^2-4ac\\\\if\ \Delta > 0\ then\ two\ solutions\\\\x_1=\dfrac{-b-\sqrt\Delta}{2a};\ \dfrac{-b+\sqrt\Delta}{2a}\\\\if\ \Delta=0\ the\ one\ solution\\\\x_o=\dfrac{-b}{2a}\\\\if\ \Delta < 0\ then\ no\ solutions.[/tex]
We have:[tex]3x^2+4x-2=0\\\\a=3;\ b=4;\ c=-2[/tex]
sustitute:[tex]\Delta=4^2-4\cdot3\cdot(-2)=16+24=40\\\\\sqrt\Delta=\sqrt{40}=\sqrt{4\cdot10}=\sqrt4\cdot\sqrt{10}=2\sqrt{10} > 0[/tex]
[tex]x_1=\dfrac{-4-2\sqrt{10}}{2\cdot3}=\dfrac{-2-\sqrt{10}}{3}\\\\x_2=\dfrac{-4+2\sqrt{10}}{2\cdot3}=\dfrac{-2+\sqrt{10}}{3}[/tex]
