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 $x_2=-0.586$

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