Question 2297
y={{{x^2-4x+2}}}

Solve this means put y=0, so we get

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

Substitution method??

You either factorise the quadratic or use the formula. Here, use the formula, as i cannot factorise it.

x = {{{(-(-4) +- sqrt((-4)^2 - 4*(1)*(2)))/2}}}

x = {{{(4 +- sqrt(16 - 8))/2}}}

x = {{{(4 +- sqrt(8))/2}}}

x = {{{(4 +- sqrt(4*2))/2}}}

x = {{{(4 +- sqrt(4)sqrt(2))/2}}}

x = {{{(4 +- 2sqrt(2))/2}}}

x = {{{2 +- sqrt(2)}}}

these are your 2 answers, in an exact form.

Use your calculator to work the approx answers out, say to 2 decimal places.

jon