kk5032019 2021-4-17 16:20
一元二次方程解，解 3(x^2) + 5x - 14 = 0 ?
sorry 之前打錯左 !
[[i] 本帖最後由 kk5032019 於 2021-4-18 06:15 PM 編輯 [/i]]
top11 2021-4-17 18:34
[quote]原帖由 [i]kk5032019[/i] 於 2021-4-17 04:20 PM 發表 [url=https://www.discuss.com.hk/redirect.php?goto=findpost&pid=534254809&ptid=29832265][img]https://www.discuss.com.hk/images/common/back.gif[/img][/url]
一元二次方程解，解 3(x^2) - 5x - 14 = 0 ? [/quote]
用公式：
ax² + bx + c = 0 ⇒ ｘ ＝ [ -b ± √(b² - 4ac) ] / (2a)
3x² - 5x - 14 = 0
x = { -(-5) ± √[ (-5)² - 4 × 3 × (-14) ] } / (2 × 3)
x = [ 5 ± √(25+168) ] /6
x = [ 5 ± √193 ] /6
kk5032019 2021-4-18 18:14
sorry 打錯左，應該係 3(x^2) + 5x - 14 = 0 ?