How do you solve the quadratic using the quadratic formula given #x^2+4x+2=0#?

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Answer 1 · 2016-11-16T11:51:40

The solutions are #S={-3.414,-0.586}#

The quadratic equation is #x^2+4x+2=0#

Let's calculate the discriminant

#Delta=b^2-4a*c=4^2-4*2=16-8=8#

As #Delta>0#

Therefore, we expect 2 real roots.

The solutions are #x=(-b+-sqrtDelta)/(2a)#

#=(-4+-sqrt8)/2=(-4+-2sqrt2)/2=-2+-sqrt2#

#x_1=-2-sqrt2# and #x_2=-2+sqrt2#

#x_1=-3.414#

and #x_2=-0.586#