Question 104861
{{{2x^2+4x-2 =0}}}
{{{x^2+2x-1=0}}}


By the quadratic formula,


{{{x=(-1+-sqrt(2^2-4*1*(-1)))/(2*1)}}}
{{{x=(-1+-sqrt(4+4))/(2)}}}
{{{x=(-1+- sqrt(2* sqrt(2)))/(2)}}}
These are the roots.


Look at my lesson entitled,"Shortcut in finding the vertex of any parabola"

Differentiating y,
{{{dy/dx=2x+2}}}
{{{0=2x+2}}}
{{{-2=2x}}}
{{{x=-1}}}

Solving for y,

{{{2x^2+4x-2 =0}}}
{{{2-4-2=y}}}
{{{y=-4}}}

Therefore, the VERTEX Is at (-1,-4)


Power up,
HyperBrain!