How do you find the zeros, real and imaginary, of #y=-4x^2-2x+3 #using the quadratic formula?

1 Answer
Feb 19, 2016

Answer:

#(-1+--1sqrt(13))/4#

Explanation:

The quadratic formula is #-b/(2a)+-sqrt(b^2-4*a*c)/(2a)#

Let's look at our equation, #-4x^2-2x+3#, in a more general form: #ax^2-bx+c#.

So #a# is #-4#
#b# is #-2#
#c# is #3#.

Now let's plug our variables into the quadratic formula. I like to put parentheses around the variables I'm replacing, just to be extra careful about silly little sign errors.

#(-(-2))/(2(-4))+-sqrt((-2)^2-4*(-4)*(3))/(2(-4))#.

This becomes #2/-8+-sqrt(4-(-48))/-8#, which we can simplify to #-1/4+-(2sqrt(13))/-8#, or #-1/4+-(-1sqrt(13))/4#.

The final answer is #(-1+--1sqrt(13))/4#