2007-06-09 14:07:37 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
-2x^2 - 4x +6 = 0
2x^2 + 4x - 6 = 0
2(x^2 + 2x - 3) = 0
x^2 + 2x - 3 =0
x^2 - x + 3x - 3 = 0
x ( x - 1 ) + 3 ( x- 1 ) = 0
(x - 1 )(x + 3 ) = 0
Therefore, x - 1 = 0 (or) x + 3 = 0
Thus, x = 1 or -3
2007-06-09 15:34:23 · answer #1 · answered by chan_l_u 2 · 0⤊ 0⤋
2x² + 4x - 6 = 0
x² + 2x - 3 = 0
x = [ - 2 ± â(4 + 12) ] / 2
x = [ - 2 ± 4 ] / 2
x = 1 , x = - 3
Check
x² + 2x - 3 = 0
(x + 3).(x - 1) = 0
x = 1 , x = - 3 (as above)
2007-06-10 10:23:15 · answer #2 · answered by Como 7 · 0⤊ 0⤋
-2x^2 - 4x + 6=0
x^2 + 2x - 3 = 0
x = [-2 +- sqrt((2)^2 - 4(1)(-3)] / 2(1)
x = [-2 +- 4] / 2
x = -3, 1
2007-06-09 21:11:52 · answer #3 · answered by Kemmy 6 · 1⤊ 0⤋
quadratic formula: y= (-b (+or-) square root of (b^2 - 4ac))/2a
plug in a,b, and c:
y= (-(-4) (+or-) square root of ((-4)^2 - 4(-2)(6)))/2(-2)
y= (4 (+or-) square root of (16 + 48)) / (-4)
y= (4 (+or-) square root of (64)) / (-4)
y= (4 (+or-) 8) / (-4)
y= (4 + 8) / (-4) = 12 / -4 = -3 and y= (4 - 8) / (-4) = -4 / -4 =1
y = -3 and y = 1
2007-06-09 21:21:54 · answer #4 · answered by Cassandra H 2 · 0⤊ 0⤋
+1, -3
a= 2, b= 4, c= -6 = plug into calculator and push "GO"
Or read what Kemmy said. d
2007-06-09 21:13:50 · answer #5 · answered by davec996 4 · 0⤊ 0⤋
(-2x-2) (x-3)=0
-2x-2 = 0
-2x = 2
x = 2/-2
x = -1
x-3 = 0
x = 3
x = -1,3
2007-06-09 21:21:45 · answer #6 · answered by Anonymous · 0⤊ 1⤋