2006-08-24 23:57:03 · 7 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
If ax^2+bx+c =0
Delta = (b^2 ) - 4(ac)
and Delta >= 0
x1 = (-b + √Delta)/2a
x2 = (-b - √Delta)/2a
2006-08-25 00:33:14 · answer #1 · answered by sweetie 5 · 1⤊ 0⤋
I'm guessing that you mean ax²+bx+c = 0 so then
x = (-b ± â(b²-4ac))/2a
You get this by 'completing the square' (which you really, **really** should study about in your math book âº)
Doug
2006-08-25 00:04:36 · answer #2 · answered by doug_donaghue 7 · 0⤊ 0⤋
x = (-b + â(b²-4ac))/2a OR
x = x = (-b - â(b²-4ac))/2a
These are known as the roots of the quadratic equation
ax^2+bx+c = 0.
Without the "=0" part, as in your question, it doesn't have an answer.
2006-08-25 00:12:00 · answer #3 · answered by blind_chameleon 5 · 0⤊ 0⤋
depends on what method will you use to solve..
wait...is it
ax^2+bx+c=0 ?
2006-08-25 00:43:11 · answer #4 · answered by L 2 · 0⤊ 0⤋
--- ax^2+bx+c = 0
--- ax^2+bx = -c
--- x^2+(b/a)x = -(c/a)
--- x^2+(b/a)x+(b^2/4a^2) = -(c/a)+(b^2/4a^2)
--- (x+b/2a)^2 = -(c/a)+(b^2/4a^2)
--- (x+b/2a)^2 = -(4ac/4a^2)+(b^2/4a^2)
--- (x+b/2a)^2 = (b^2-4ac)/4a^2
--- |x+b/2a| = sqrt(b^2-4ac)/|2a|
--- x+b/2a = +-sqrt(b^2-4ac)/2a
--- x = -b/2a +- sqrt(b^2-4ac)/2a
--- x = [-b +- sqrt(b^2-4ac)]/2a
2006-08-25 01:14:23 · answer #5 · answered by Kyrix 6 · 0⤊ 0⤋
gee.....your math book really sux! or you....nmd
Doug is right!
2006-08-25 03:14:26 · answer #6 · answered by David F 2 · 0⤊ 0⤋
x = (-b ± â(b²-4ac))/2a
2006-08-25 00:42:58 · answer #7 · answered by Anonymous · 0⤊ 0⤋