SOLUTION: Find the real or imaginary solutions by using the quadratic formula. 2x^2+4x=-1

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Question 869811: Find the real or imaginary solutions by using the quadratic formula. 2x^2+4x=-1
Found 3 solutions by mananth, Fombitz, ewatrrr:
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
2x%5E2%2B4x%2B1=0
compare with ax%5E2%2Bbx%2Bc=0
a= 2 , b= 4 , c= 1

b^2-4ac= 16 + -8
b^2-4ac= 8
%09sqrt%28%098%09%29=%092.83%09
x=%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F%282a%29
x1=%28-b%2Bsqrt%28b%5E2-4ac%29%29%2F%282a%29
x1=( -4 + 2.83 )/ 4
x1= -0.29
x2=%28-b-sqrt%28b%5E2-4ac%29%29%2F%282a%29
x2=( -4 -2.83 ) / 4
x2= -1.71

Answer by Fombitz(32388) About Me  (Show Source):
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!


Hi

2x^2+4x + 1 = 0     ax^2 + x + c  Format

x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+

x+=+%28-4+%2B-+sqrt%28+8+%29%29%2F4+

x+=+%28-4+%2B-+sqrt%28+4%2A2+%29%29%2F4+

x+=+%28-4+%2B-+2sqrt%282+%29%29%2F4+

 x = -1 ± .5√2