How do you solve $2 x^{2} + 8 x - 3 = 0$ $2x^2 + 8x - 3 = 0$ ?

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How do you solve $2 x^{2} + 8 x - 3 = 0$ $2x^2 + 8x - 3 = 0$ ?

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Answer 1 · 2016-04-04T05:06:21

$x = - 2 \pm \frac{\sqrt{22}}{2}$ $x = - 2 +- sqrt22/2$

Explanation:

Use the improved quadratic formula (Google, Yahoo Search)
$D = d^{2} = b^{2} - 4 a c = 64 + 24 = 88$ $D = d^2 = b^2 - 4ac = 64 + 24 = 88$ ---> $d = \pm 2 \sqrt{22}$ $d = +-2sqrt22$
There are 2 real roots:
$x = - \frac{b}{2 a} \pm \frac{d}{2 a} = - \frac{8}{4} \pm \frac{2 \sqrt{22}}{4} =$ $x = -b/(2a) +- d/(2a) = -8/4 +- (2sqrt22)/4 =$
$= - 2 \pm \frac{\sqrt{22}}{2}$ $= -2 +- sqrt22/2$

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How do you solve $2 x^{2} + 8 x - 3 = 0$ $2x^2 + 8x - 3 = 0$ ?

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