Question 417052
It's not nearly as complicated as you make it seem.
The formula is:
{{{x = (-b+-sqrt(b^2-4ac))/2a}}}
To solve for x in: {{{3x^2+4x-8 = 0}}} Substitute into the formula: a = 3, b = 4, and c = -8 as you did already.
{{{x = (-4+-sqrt(4^2-4(3)(-8)))/2(3)}}} Now simplify his lot.
{{{x = (-4+-sqrt(16-(-96)))/6}}}
{{{x = (-4+-sqrt(112))/6}}} Now simplify the square root.
{{{x = (-4+-sqrt(16*7))/6}}} {{{sqrt(16) = 4}}} so put the 4 in front of the radical leaving the 7 inside.
{{{x = (-4/6)+(4/6)sqrt(7)}}} and {{{x = (-4/6)-(4/6)sqrt(7)}}} Simplify the fractions.
{{{x = (-2+2sqrt(7))/3}}} and {{{(-2-2sqrt(7))/3}}}