Solve for X x^2+6x+4=0
2006-11-09 10:51:47 · 8 answers · asked by pfcmitchell1987 2 in Science & Mathematics ➔ Mathematics
delta = 6^2 - 4*4 = 20 ; square root of delta = 2 * square root of 5
x1 = (-6 + 2 square root of 5)/2 = square root of 5 - 3
x2 = (-6 - 2 square root of 5)/2 = -(square root of 5 + 3)
2006-11-09 10:55:48 · answer #1 · answered by James Chan 4 · 0⤊ 0⤋
OK x^2 - 6x -4 = 0 x^2 -6x -4 +13 = 13 x^2 -6x +9 = 13 (x-3)^2 = 13 So the answer is 3 +- sqrt 13. Hope that helps.
2016-05-22 01:27:33 · answer #2 · answered by Anonymous · 0⤊ 0⤋
x^2 + 6x + 4 = 0
x = (-b ± sqrt(b^2 - 4ac))/(2a)
x = (-6 ± sqrt(6^2 - 4(1)(4)))/(2(1))
x = (-6 ± sqrt(36 - 16))/2
x = (-6 ± sqrt(20))/2
x = (-6 ± sqrt(4 * 5))/2
x = (-6 ± 2sqrt(5))/2
x = -3 ± sqrt(5)
ANS : x = -3 + sqrt(5) or -3 - sqrt(5)
2006-11-09 14:49:32 · answer #3 · answered by Sherman81 6 · 0⤊ 0⤋
X= -6+or-(sqrt of [6*6] -4*4) so x= -6+2*SQRT(5) OR
-6-2*SQRT(5)
2006-11-09 12:34:49 · answer #4 · answered by Anonymous · 0⤊ 0⤋
x^2 + 6x = -4
8x = -4
x = -1/2
2006-11-09 10:55:03 · answer #5 · answered by Chris C 3 · 0⤊ 0⤋
its very simple...just use the quadratic formula...
x=(-b+(b^2-4ac)^1/2)/2a
x=(-b-(b^2-4ac)^1/2)/2a
pluging in the a,b,and c given ax^2+bx+c
you get
x=-3+(5)^1/2
x=-3-(5)^1/2
2006-11-09 10:58:35 · answer #6 · answered by Anonymous · 0⤊ 0⤋
x^2+6x+4=0
x=(-6+/-√(36-4*4*1))/2
x=-3+/-(1/2)√20
x=-3+/-√5
x=-0.7639
x=-5.236
2006-11-09 11:01:40 · answer #7 · answered by yupchagee 7 · 0⤊ 0⤋
x=-5.23607
2006-11-09 11:02:37 · answer #8 · answered by kuemerle5 1 · 0⤊ 0⤋