Question 327388
Temporarily substitute {{{y = x^2}}} so now you have:
{{{2y^2-10y = -8}}} Add 8 to both sides.
{{{2y^2-10y +8 = 0}}} So now you have a quadratic equation in y. Factor a 2.
{{{2(y^2-5y+4) = 0}}} so...
{{{y^2-5y+4 = 0}}} This will factor nicely into...
{{{(y-1)(y-4) = 0}}} and from the zero product rule you get...
{{{y-1 = 0}}} or {{{y-4 = 0}}} therefore...
{{{y = 1}}} or {{{y = 4}}} Re-substtute {{{y = x^2}}}
{{{x^2 = 1}}} or {{{x^2 = 4}}} Taking the square root of both of these, we get...
{{{x = 1}}} or {{{x = -1}}} or {{{x = 2}}} or {{{x = -2}}}