How do you solve #2x^2 + 4x = -3#?

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Answer 1 · 2016-12-02T05:43:03

#x=-1+-isqrt(-1)/2#

Given -

#2x^2+4x=-3#

Divide by 2

#(2x^2)/2+(4x)/2=(-3)/2#

#x^2+2x=(-3)/2#

take half of the coefficient of #x# square it and add on both sides

#x^2+2x+1=(-3)/2+1#
#(x+1)^2=(-3+2)/2=(-1)/2#

Taking square on both sides

#x+1=sqrt(-1)/2#

#x=-1+-isqrt(-1)/2#

Answer 2 · 2016-12-02T06:09:39

#x=-1+(sqrt2i)/2#, #x=-1-(sqrt2i)/2#

Solve: #2x^2+4x=-3#

Add #3# to both sides.

#2x^2+4x+3=0# is a quadratic equation, #ax^2+bx+c#, where #a=2#, #b=4#, and #c=3#. The quadratic formula can be used to solve this equation.

Quadratic Formula

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

Substitute the given values into the formula and solve for #x#.

#x=(-4+-sqrt(4^2-(4*2*3)))/(2*2)#

#x=(-4+-sqrt(16-24))/4#

#x=(-4+-sqrt(-8))/4#

Simplify.

#x=(-4+-2sqrt2i)/4#

Simplify.

#x=-1+(sqrt2i)/2#, #x=-1-(sqrt2i)/2#