Question 256204
Solving 1 at a time, we get:
1. 2x^2 - 3x - 11 = 0
we can't factor, so we use the quadratic as
{{{x = (3 +- sqrt(9-4*2*(-11)))/(4) }}}
and we get
{{{x = (3+- sqrt(91)))/4}}}
--
2. 3x^2 + 48 = 0
divide by 3 to get
{{{x^2+16=0}}}
subtract 16 to get
{{{x^2 = -16}}}
take a square root and we get
no solution
OR
+ or - 4i.
3. 2x^2 - x + 1 = 0
we can't factor this so we apply quadratic as
{{{x = (1 +- sqrt(1-4*2*1))/(4)}}}
and we get
{{{x = (1 +- sqrt(-7))/(4)}}}
which has no solution or has imaginary roots.