Find the solutions of the quadratic equation 14x^2+9x+10=014x
2
+9x+10=014, x, start su...
Mathematics, 28.01.2020 13:35 emurray
Find the solutions of the quadratic equation 14x^2+9x+10=014x
2
+9x+10=014, x, start superscript, 2, end superscript, plus, 9, x, plus, 10, equals, 0.
choose 1
choose 1
(choice a)
a
\dfrac{9}{28}\pm\dfrac{\sqrt{479}}{ 28}i
28
9
±
28
479
istart fraction, 9, divided by, 28, end fraction, plus minus, start fraction, square root of, 479, end square root, divided by, 28, end fraction, i
(choice b)
b
-\dfrac{9}{28}\pm\dfrac{\sqrt{479}} {28}i−
28
9
±
28
479
iminus, start fraction, 9, divided by, 28, end fraction, plus minus, start fraction, square root of, 479, end square root, divided by, 28, end fraction, i
(choice c)
c
-\dfrac{9}{28}\pm\dfrac{\sqrt{479}} {28}−
28
9
±
28
479
minus, start fraction, 9, divided by, 28, end fraction, plus minus, start fraction, square root of, 479, end square root, divided by, 28, end fraction
(choice d)
d
\dfrac{9}{28}\pm\dfrac{\sqrt{479}}{ 28}
28
9
±
28
479
Answers: 1
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