Help with solving an equation?

Question

If 2x^2+4x=-3 is being solved, what is the value of x? Showing all steps to solve this will be very helpful. Thank you.

Answer 1

After completing this tutorial, you will be a master at solving quadratic equations.
http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut17_quad.htm
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Answer 2

by factoring

1/2 (2x^2 + 4x +3 = 0)
x^2 + 2x + 3/2 = 0
x^2 + 2x + 1 = -1/2
(x + 1)^2 = -1/2

can't square root negatives without complex numbers...
by formula

-4 + (16 - [2 x 3 x 4])^1/2 / 8

b^2- 4ac < 0 so there isnt a solution

Answer 3

2x²+4x=-3
2x² + 4x + 3 = 0
delta = b² - 4ac
d = 4² - 4*2*3
d = 16 - 24
d = -8
d < 0 => There is no Real root.
>;;<

Answer 4

2x^2+4x=-3
2x^2+4x+3=0
quadratic formula:
x=(-b+or-sqrt(b^2-4*a*c))/2a
for every quadratic equation of the form ax^2 + bx + c=0
x=(-4+or-sqrt(4^2-4(2)(3))/2(2)
=(-4+or-sqrt(-8))/4
= -4+(2i)(sqrt(2))/4 or -4-(2i)(sqrt(2))/4
= (-2 + or - (i)(sqrt(2))

Answer 5

Guess till you get it...
(2 * ((-.7)^2)) + (4 * (-.7)) = -1.82
(2 * ((-.9)^2)) + (4 * (-.9)) = -1.98
(2 * ((-1.9)^2)) + (4 * (-1.9)) = -0.38
(2 * ((-1.1)^2)) + (4 * (-1.1)) = -1.98
(2 * ((-1.01)^2)) + (4 * (-1.01)) = -1.9998
I think 4x + 4 is the derivative, so
x = -1 makes the horizontal tangent,
the minimum value of -2.
I don't think you can get to -3.