solve by using the quadratic formula
2007-05-15 04:49:42 · 8 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
3x^2+5x+2=0
a=3 b = 5 c = 2
x = [-b +/- sqrt(b^2-4ac) ]/2a
x = [-5 +/- sqrt(25-24)] /6
x = [-5 +/-1 ] / 6
x = -5 + 1 /6, x = -5 -1 /6
x=-2/3, x = -1
2007-05-15 04:54:24 · answer #1 · answered by builder-mech 2 · 0⤊ 1⤋
First, we put the quadratic equation together on one side = 0. 3x^2+5x+2=0. We identify our coefficients: a=3, b=5,c=2. Then we substitute in quad formula solution eqtn:
y = [- 5 +/- sqrt (1)]/6 or y = -1 or -2/3
2007-05-15 04:55:22 · answer #2 · answered by cattbarf 7 · 0⤊ 1⤋
3x^2= -5x-2
3x^2+5x+2=0
a=3
b=5
c=2
plug these into the quadratic formula and solve.
2007-05-15 04:53:52 · answer #3 · answered by jaybee 4 · 0⤊ 1⤋
3x² + 5x + 2 = 0
x = [ - 5 ± √(25 - 24) ]/ 6
x = [ - 5 ± √(1) ] / 6
x = - 2/3 , x = - 1
2007-05-15 07:53:29 · answer #4 · answered by Como 7 · 0⤊ 1⤋
3x^2 + 5x +2 = 0
x = (-5 +- sqrt(5^2 - 4*3*2))/2*3
= (-5 +- sqrt (1) )/ 6 = -1, -2/3
2007-05-15 04:53:58 · answer #5 · answered by gesges 3 · 0⤊ 1⤋
3x² = -5x - 2
3x² + 5x + 2 = 0
ax² +bx+c = 0
a= 3
b= 5
c = 2
. . . .-b±√(b²-4ac)
x = -------------------
... . . . . . .2a
. . . .- 5±√(25-4*3*2)
x = ---------------------
. . . . . . . . 2*3
x = (-5±1)/6
x = -1
x = -4/6 = - 2/3
ANSWER: x = -1, x = - 2/3
2007-05-15 04:55:14 · answer #6 · answered by zindagii_peru 4 · 1⤊ 0⤋
I would help you but I suck at math.
2007-05-15 04:57:44 · answer #7 · answered by Alex 3 · 0⤊ 1⤋
dont know
2007-05-15 04:58:02 · answer #8 · answered by David 2 · 0⤊ 1⤋