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1 year ago

9

Solve quadratic equation 10x^2+x-2=0

1 answer:

zhuklara [117]1 year ago

8 0

Answer:

$\{-\frac{1}{2} \} \cup \{\frac{2}{5}\}$

Step-by-step explanation:

$10x^2 + x - 2 = 0; \\ a = 10, b = 1, c = -2; \\ D = b^2 - 4ac = 1^2 - 4 * 10 * (-2) = 1 + 80 = 81 = 9^2, > 0; \\ x_{1, 2} = \frac{-b \pm \sqrt{D}}{2a} = \frac{-1 \pm \sqrt{9^2}}{2 * 10} = \frac{-1 \pm 9}{20} = \left \ [ {{\frac{-1 - 9}{20} = -\frac{10}{20} = -\frac{1}{2} } \atop {\frac{-1 + 9}{20} = \frac{8}{20} = \frac{2}{5}}} \right.$

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2 years ago

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Answer:

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2 years ago

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Answer:

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Step-by-step explanation:

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2 years ago

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Step-by-step explanation:

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Hope This Helps :)

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2 years ago