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Bas_tet [7]
3 years ago
10

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

Mathematics
2 answers:
DENIUS [597]3 years ago
7 0
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}



Step2247 [10]3 years ago
5 0

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

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

<span>x =  \frac{8 +/-  \sqrt{64 - 24}}{6}</span>

<span>x =  \frac{8 +/-  \sqrt{40}}{6}</span>

<span>x = \frac{8 +/- 6.31}{6}</span>

<span>x = \frac{14.32}{6}          x ≈ 2.39</span>

<span>x = \frac{1.68}{6}            x ≈ 0.28</span>

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