Question 177528
The solutions to a quadratic in the form {{{ax^2 + bx + c = 0}}} are given by:


{{{x = (-b +- sqrt( b^2-4ac ))/(2a) }}} 


For your problem:  {{{a = 6}}}, {{{b = 3}}}, and {{{c = -18}}}, but these numbers have a common factor that you can eliminate, namely 3.  Hence, {{{a = 2}}}, {{{b = 1}}}, and {{{c = -6}}}, so:


{{{x = (-1 +- sqrt( 1^2-4(2)(-6)))/(2*2) }}}.  Now just do the arithmetic.



I'm pretty sure you mean {{{y = x^2 + 4x - 5}}} instead of what you wrote.


{{{drawing(
400, 400, -6, 6, -6, 6,
grid(1),
graph(
400, 400, -6, 6, -6, 6,
y = x^2 + 4x - 5
))}}}


The discriminant is just the part of the quadratic formula that is under the radical, {{{DELTA = b^2 - 4ac}}}, so calculate {{{DELTA = 4^2 - 4(1)(4)}}}