quadratic equation: 2x-3x^2-4=0?

this page shows your exact problem worked out....

http://www.mathway.com/answer.aspx?p=calg?p=2x-3x&CC&2-4&EE&0?p=112?p=?p=?p=?p=?p=?p=?p=0

3x² - 2x + 4 = 0
x = [ 2 ± √ (4 - 48 ) ] / 6
x = [ 2 ± √ (- 44 ) ] / 6
x = [ 2 ± √ ( 44 i² ) ] / 6
x = [ 2 ± 2√(11) i ] / 6
x = [ 1 ± √(11) i ] / 3

the ecuation has no real solution

2x - 3x^2 - 4 = 0

3x^2 - 2x + 4 = 0

ac = 3(4) = 12

There are no factors of +12 that add to -2; use quadratic formula

x = [-b +/- sqrt(b^2 - 4ac)]/2a

x = [2 +/- sqrt(4 - 48)]/6

simplify from here

x1 = (1/3) + sqrt (-11) /3
x2 = (1/3) - sqrt (-11) /3

2x - 3x^2 - 4 = 0
-3x^2 + 2x - 4 = 0
x = [-b ±√(b^2 - 4ac)]/2a

a = -3
b = 2
c = -4

x = [-2 ±√(4 - 48)]/-6
x = [-2 ±√-44]/-6 (imaginary number)
(no real roots)

r u kiddin?

3x2-2x+4=0

using quadratic formula:
2+-sqrt (4-44)

as the b^2 -4ac is -ve the eqn has no real roots!


OR u have posted the wrong eqn!