Answer :
Calculating delta:
Δ=b²-4ac
a=3
b=-6
c=-2
Δ=36-4*3*(-2)=36+24=60
√Δ=√60
Delta is positive so there are two roots:
x1=[tex] \frac{ -b+ \sqrt[2]{delta} }{2a} [/tex]
x1=[tex] \frac{6+ \sqrt{delta} }{2*3} [/tex]=[tex] \frac{3+ \sqrt{15} }{3} [/tex]
x2=[tex] \frac{3- \sqrt{15} }{3} [/tex]
Δ=b²-4ac
a=3
b=-6
c=-2
Δ=36-4*3*(-2)=36+24=60
√Δ=√60
Delta is positive so there are two roots:
x1=[tex] \frac{ -b+ \sqrt[2]{delta} }{2a} [/tex]
x1=[tex] \frac{6+ \sqrt{delta} }{2*3} [/tex]=[tex] \frac{3+ \sqrt{15} }{3} [/tex]
x2=[tex] \frac{3- \sqrt{15} }{3} [/tex]
[tex]3x^2-6x-2=0\\\\(\sqrt3\ x)^2-2\cdot\sqrt3\ x\cdot\sqrt3+(\sqrt3)^2-(\sqrt3)^2-2=0\\\\(\sqrt3\ x-\sqrt3)^2-3-2=0\\\\(\sqrt3\ x-\sqrt3)^2=5\iff\sqrt3\ x-\sqrt3=-\sqrt5\ \vee\ \sqrt3\ x-\sqrt3=\sqrt5\\\\\sqrt3\ x=\sqrt3-\sqrt5\ \vee\ \sqrt3\ x=\sqrt3+\sqrt5\ \ \ \ \ |multiply\ both\ sides\ by\ \sqrt3\\\\3x=3-\sqrt{15}\ \vee\ 3x=3+\sqrt{15}\ \ \ \ \ \ \ |divide\ both\ sides\ by\ 3\\\\x=\frac{3-\sqrt{15}}{3}\ \vee\ x=\frac{3+\sqrt{15}}{3}[/tex]
[tex]Prove:\\\\3x^2-6x-2=0\\a=3;\ b=-6;\ c=-2\\\Delta=b^2-4ac\to\Delta=(-6)^2-4\cdot3\cdot(-2)=36+24=60\\\\x_1=\frac{-b-\sqrt\Delta}{2a};\ x_2=\frac{-b+\sqrt\Delta}{2a}\\\\\sqrt\Delta=\sqrt{60}=\sqrt{4\cdot15}=\sqrt4\cdot\sqrt{15}=2\sqrt{15}\\\\x_1=\frac{6-2\sqrt{15}}{2\cdot3}=\frac{3-\sqrt{15}}{3}\ \vee\ x_2=\frac{6+2\sqrt{15}}{2\cdot3}=\frac{3+\sqrt{15}}{3}[/tex]
[tex]Prove:\\\\3x^2-6x-2=0\\a=3;\ b=-6;\ c=-2\\\Delta=b^2-4ac\to\Delta=(-6)^2-4\cdot3\cdot(-2)=36+24=60\\\\x_1=\frac{-b-\sqrt\Delta}{2a};\ x_2=\frac{-b+\sqrt\Delta}{2a}\\\\\sqrt\Delta=\sqrt{60}=\sqrt{4\cdot15}=\sqrt4\cdot\sqrt{15}=2\sqrt{15}\\\\x_1=\frac{6-2\sqrt{15}}{2\cdot3}=\frac{3-\sqrt{15}}{3}\ \vee\ x_2=\frac{6+2\sqrt{15}}{2\cdot3}=\frac{3+\sqrt{15}}{3}[/tex]
