Solve the following quadratic $ 21x^{2}-28x+10 = 0 $
- $ \dfrac{2}{3}\pm \dfrac{\sqrt{15}}{21}i $
- $ \dfrac{2}{3}\pm \dfrac{\sqrt{16}}{21}i $
- $ \dfrac{2}{3}\pm \dfrac{\sqrt{14}}{21}i $
- $ \dfrac{-2}{3}\pm \dfrac{\sqrt{14}}{21}i $
Reveal answer
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C
Correct answer
Explanation
Using the quadratic formula x = (-b +/- sqrt(b^2 - 4ac)) / 2a, we have a=21, b=-28, c=10. The discriminant is (-28)^2 - 4(21)(10) = 784 - 840 = -56. The roots are (28 +/- sqrt(-56)) / 42 = (28 +/- i*sqrt(56)) / 42 = (28 +/- 2*i*sqrt(14)) / 42 = 2/3 +/- i*sqrt(14)/21.
AI explanation
Using the quadratic formula x = (-b plus or minus sqrt(b^2 - 4ac)) / 2a, we substitute a = 21, b = -28, and c = 10. The discriminant is (-28)^2 - 4(21)(10) = 784 - 840 = -56. The roots are (28 plus or minus sqrt(-56)) / 42, which simplifies to 2/3 plus or minus (i*sqrt(14))/21.