Question 168902
{{{x^2-4x+2=0}}}
Remember ----> {{{system(a=1,b=-4,c=2)}}}
Via Pyth. Theorem:
{{{x=(-b+-sqrt(b^2-4ac))/(2a)}}}
{{{x=(-(-4)+-sqrt(-4^2-4*1*2))/(2*1)}}}
{{{x=(4+-sqrt(16-8))/2=(4+-sqrt(8))/2=(4+-2.8284)/2}}}
2 values:
{{{x=(4+2.8284)/2=6.8284/2=highlight(3.4142)}}}
{{{x=(4-2.8284)/2=2.8284/2=highlight(1.4142)}}}
Let's check, use {{{x=3.4142}}}
{{{3.4142^2-4(3.4142)+2=0}}}
{{{11.65676-13.6568+2=0}}}
{{{-2.00004+2=0}}}
{{{0.00004=0}}}, close enough amigo (rounding off)
Thank you,
Jojo