this is out of chapter 10 section 7 how do u do it using ax^2+bx+c
2007-02-05 14:08:09 · 5 answers · asked by Shante R 1 in Science & Mathematics ➔ Mathematics
how to solve it.it called using quadratic forumals using ax^2+bx+c
2007-02-05 14:13:06 · update #1
your answers are -2.20 and -3.03 or 2.20 and -3.03 or 2.20 and 3.03 or -2.20 and 3.03
2007-02-05 14:18:22 · update #2
so whats da soultion
2007-02-05 14:26:00 · update #3
i still dont get how u got that answer
2007-02-05 14:32:55 · update #4
Use the quadratic Formula
-b(+-) sqrt(b^2-4ac)
-------------------------
2a
2007-02-05 14:13:20 · answer #1 · answered by Aaron 4 · 0⤊ 0⤋
6x^2 + 5x - 40 = 0
Seeing as this is of the form ax^2 + bx + c = 0, we can use the quadratic formula.
x = [-b +/- sqrt(b^2 - 4ac)] / (2a)
Plug in a = 6, b = 5, and c = -40.
x = [-5 +/- sqrt(25 - 4(6)(-40))] / (2(6))
x = [-5 +/- sqrt(25 + 960)] / 12
x = [-5 +/- sqrt(985)] / 12
So your two answers are:
x = [-5 + sqrt(985)] / 12
x = [-5 - sqrt(985)] / 12
2007-02-05 22:16:30 · answer #2 · answered by Puggy 7 · 0⤊ 0⤋
a = 6, b = 5, c = -40
2007-02-05 22:10:51 · answer #3 · answered by kippie2525 3 · 0⤊ 0⤋
Use the quadratic formula
2007-02-05 22:15:55 · answer #4 · answered by Tom K 1 · 0⤊ 0⤋
a = 6, b = 5, c = -40
x = (-b ±â(b²-4ac))/(2a)
It's just plug 'n play from there âº
Doug
2007-02-05 22:13:20 · answer #5 · answered by doug_donaghue 7 · 0⤊ 0⤋