How do you solve $3 x^{2} - 5 x + 4 = 0$ $3x^2-5x+4=0$ using the quadratic formula?

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How do you solve $3 x^{2} - 5 x + 4 = 0$ $3x^2-5x+4=0$ using the quadratic formula?

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Answer 1 · 2016-05-11T03:38:43

$x = \frac{5 \pm i \sqrt{23}}{6}$ $x = (5 +- isqrt23)/6$

Explanation:

$y = 3 x^{2} - 5 x + 4 = 0$ $y = 3x^2 - 5x + 4 = 0$
Use the improved quadratic formula.
$D = d^{2} = b^{2} - 4 a c = 25 - 48 = - 23 = 23 i^{2}$ $D = d^2 = b^2 - 4ac = 25 - 48 = - 23 = 23i^2$ --> $d = \pm i \sqrt{23}$ $d = +- isqrt23$
There are no real roots. There are 2 complex roots:
$x = - \frac{b}{2 a} \pm \frac{d}{2 a} = \frac{5}{6} \pm i \frac{\sqrt{23}}{6} = \frac{5 \pm i \sqrt{23}}{6}$ $x = -b/(2a) +- d/(2a) = 5/6 +- isqrt23/6 = (5 +- isqrt23)/6$

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How do you solve $3 x^{2} - 5 x + 4 = 0$ $3x^2-5x+4=0$ using the quadratic formula?

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