The positive root of t in the following equation, 4t2 + 7t - 1 = 0, correct to 4 places of decimal, is
\(4t^{2} + 7t - 1 = 0\)
Using a = 4, b = 7, c = -1.
\(x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\)
= \(\frac{-7 \pm \sqrt{7^{2} - 4(4)(-1)}}{2(4)}\)
= \(\frac{-7 \pm \sqrt{49 + 16}}{8}\)
= \(\frac{-7 \pm \sqrt{65}}{8}\)
= \(\frac{-7 \pm 8.0623}{8}\)
The positive answer = \(\frac{-7 + 8.0623}{8} = \frac{1.0623}{8}\)
\(\approxeq 0.1328\) (4 decimal place)
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