How do i solve 3x² - 8x + 2 = 0 using quadratic formula?

\( x = \frac{-b +/- \sqrt{b^2 - 4ac}}{2a} \)

\( x = \frac{-(-8) +/- \sqrt{(-8)^2 - 4(3)(2)}}{2(3)} \)

\( x = \frac{8 +/- \sqrt{64 - 24}}{6} \)

\( x = \frac{8 +/- \sqrt{40}}{6} \)

\( x = \frac{8 +/- 6.31}{6} \)

\( x = \frac{14.32}{6} \)          x ≈ 2.39

\( x = \frac{1.68}{6} \)            x ≈ 0.28

\( 3x^2 - 8x + 2 = 0\\ \\a=3, \ \ b=-8, \ \ c=2 \\ \\x_{1}=\frac{-b-\sqrt{b^2-4ac}}{2a} =\frac{8-\sqrt{ (-8)^2-4 \cdot 3\cdot 2}}{2 \cdot 3} =\frac{8-\sqrt{ 64-24 }}{6} =\\ \\ =\frac{8-\sqrt{40 }}{6} = \frac{ 8-\sqrt{4\cdot 10 } }{6} = \frac{ 8-2\sqrt{ 10 }}{6} = \frac{2 (4-\sqrt{ 10 })}{6} = \frac{ 4-\sqrt{ 10 } }{3} \)

\( x_{2}=\frac{-b+\sqrt{b^2-4ac}}{2a} =\frac{8+\sqrt{ (-8)^2-4 \cdot 3\cdot 2}}{2 \cdot 3} = \frac{ 4+\sqrt{ 10 } }{3} \)




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