how to solve.3x^2+4x-1=0?

Answer 1

METHOD 1: Quadratic formula

a = 3, b = 4, c = -1

........-b ± sqrt( b² - 4ac )
x = -----------------------------
.................. 2a

........-4 ± sqrt( 4² - 4(3)(-1) )
x = -----------------------------
.................. 2(3)

........-4 ± sqrt( 16 + 12 )
x = -----------------------------
.................. 6

........-4 ± sqrt( 28 )
x = ---------------------
.................. 6

........-4 ± sqrt(4) sqrt(7)
x = ---------------------------
.................. 6

........-4 ± 2 sqrt(7)
x = --------------------
.................. 6

........-2 ± sqrt(7)
x = ------------------
................ 3

x = -(2/3) ± (1/3)sqrt(7)

METHOD 2: Completing the square:

3x² + 4x - 1 = 0
3x² + 4x = 1
x² + (4/3)x = (1/3)
x² + 2(2/3)x + (2/3)² = (1/3) + (2/3)²
(x + (2/3))² = 3/9 + 4/9
(x + (2/3))² = 7/9

x + (2/3) = ±sqrt(7/9)
x + (2/3) = ±sqrt(7) / sqrt(9)
x + (2/3) = ±sqrt(7) / 3
x + (2/3) = ±(1/3)sqrt(7)

x = -(2/3) ± (1/3)sqrt(7)

Answer 2

Is that supposed to be
(a) 0.3x² + 4x - 1 = 0
or
(b) -3x² + 4x - 1 = 0
or
(c) 3x² + 4x - 1 = 0 ?

(b) makes sense because it is easy to factor:
-3x² + 4x - 1 = (x - 1)(-3x + 1) = 0

Answer 3

a = 3
b = 4
c = -1

x = [ -b PLUSMINUS SQRT(b^2 - 4ac) ] / (2a)
x = [ -4 PLUSMINUS SQRT(4^2 - 4*3*-1) ] / (2*3)

0.215250437
-1.54858377

Answer 4

Drop back five and punt.
kidding...

the quadratic formula should do the trick..

Answer 5

use the quadratic formula
x= [- b +(or)- sqrt(b^2-4ac) ] / (2a)

Answer 6

www.webmath.com
step by step and answers too :)

Answer 7

figure it out yourself....